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Table of Contents
How do you prove a midpoint in geometry?
The Mid- Point Theorem can also be proved using triangles. Suppose two lines are drawn parallel to the x and the y-axis which begin at endpoints and connected through the midpoint, then the segment passes through the angle between them resulting in two similar triangles.
What is the relationship between a midpoint and an angle bisector?
| Term | Definition |
|---|---|
| midpoint | The midpoint of a line segment is the point on the line segment that splits the segment into two congruent parts. |
| perpendicular bisector | A segment bisector that intersects the segment at a right angle. |
What is converse midpoint theorem?
The converse of the midpoint theorem states that a line drawn parallel to a side from one side’s midpoint to the other side bisects the other side. The application of the converse of the basic proportionality theorem is the converse of the midpoint theorem.
Prove that the lines parallel to the angle bisectors through the midpoints of sides of a triangle are concurrent at the Spieker center
Last modified on 07/25/2022
Midpoint theorem and inversion of the midpoint theorem in the triangle: statement, proof, examples
Midpoint Theorem and Inverse of the Midpoint Theorem in Triangle: Geometry is the branch of mathematics that deals with different shapes and objects. Triangles are one of the most important subjects in geometry. A triangle is the smallest polygon composed of three line segments: Midpoint Theorem and Inverse Midpoint Theorem deal with the midpoints of the triangle. A midpoint is the midpoint of a line segment equidistant from either end. The midpoint theorem is also used in the fields of coordinate geometry, calculus, and algebra.
Let’s discuss the propositions, proofs of the midpoint theorem and converse of the midpoint theorem with examples in this article.
Learn the concepts of the midpoint theorem
Center point theorem in triangles
The midpoint theorem states that a line joining the midpoints of any two sides of a triangle is parallel to the other side of the triangle and has a value equal to half the value of the other parallel side.
The midpoint theorem is one of the applications of the fundamental theorem of proportionality, or Thales’ theorem, which is also used in coordinate geometry, calculus, and algebra.
Inversion of the midpoint theorem
The converse of the midpoint theorem is the statement of the midpoint theorem of the triangle. The converse of the midpoint theorem states that a line drawn parallel to one side from the midpoint of one side to the other bisects the other side.
Applying the inverse of the basic theorem of proportionality is the inverse of the midpoint theorem.
Center point theorem in the triangle proof
The midpoint theorem gives the relationship between the line connections between the midpoints of any two sides of the triangle and the other side of the triangle.
Statement: In a triangle, if a line joining the midpoints of any two sides of the triangle is parallel to the third side of the triangle and its value is equal to half the measure of the third side of the triangle.
Proof: We have to prove that \(DE\) is parallel to \(BC\) and equal to half of \(BC.\)
Proof: Let us consider the proof describing the center point theorem in triangles. Imagine a triangle (Say \(△ABC\)),
The points \(D\) and \(E\) are the midpoints of the side \(AB\) and \(AC\) of the triangle \(△ABC.\)
\(AD=DB\) and \(AE = EC \cdots \cdots {\rm{(i)}}\)
\(DE\) is the line drawn from the midpoints of the sides \(AB\) and \(AC\) of the triangle \(△ABC.\).
We have to prove that the line connecting the midpoints \(D,\) and \(E\) of the sides \(AB\) and \(AC\) of the triangle \(ABC\) is parallel to the third side \ is. (BC.\)
Extend the line \(DE\) to the point \(F,\) so that \(DE\) equals \(EF.\)
Connect the points \(C\) and \(F.\)
Here the lines \(AB\) and \(CF\) are parallel and the line \(AC\) is transverse.
We know that the pair of alternating interior angles formed by two parallel lines through a transversal are equal.
So the angles \(DAE\) and \(ECF\) are equal because they are alternating interior angles formed by the parallel lines \(AB\) and \(CF\) with transversal \(AC.\).
\(\angle DAE = \angle ECF\) (alternative interior angles)
Well, in the triangles \(ADE\) and \(ECF,\)
Sides \(AE=EC\) (Assume \(E\) is the midpoint of side \(AC\))
\(\angle AED = \angle CEF\) (vertically opposite angles)
\(\angle DAE = \angle ECF\) (alternative interior angles)
According to the A.S.A congruence rule, two triangles \(ADE\) and \(CEF,\) are congruent.
\(△ADE≅△CEF\)
\(AF=DE\) (from C.P.C.T)
And \(CF=AD\) (from C.P.C.T)
From \({\rm{(i),}}\) \(BD=CF\)
Also, by construction, \(BD\) is parallel to \(CE.\)
So \(BDFC\) is a parallelogram.
This means \(DF∥BC,\) which implies that \(DE\) is parallel to \(BC.\).
We know that \(DE = EF = \frac{1}{2}DF = \frac{1}{2}BC\)
Therefore we can say that the line connecting the midpoints (\(D\) and \(E\)) of the sides \(AB\) and \(AC\) is parallel to the other side \(BC\) and equals half of \(BC.\)
\(DE∥BC\) and \(DE = \frac{1}{2}B\)
Inverse of the Midpoint Theorem Proof
This theorem is the converse proof of the center point theorem of triangles.
Statement: The line drawn through the midpoint of any side of the triangle parallel to the other side of the triangle bisects the third side of the triangle.
Proof: We have to prove that the point \(E\) is the midpoint of the side \(AC (AE=EC).\)
Proof: Let \(D\) be the midpoint of the side \(AB\) of the triangle \(△ABC.\)
\(AD = BD \cdots \cdots {\rm{(i)}}\)
Line \(DE\) is drawn from point \(D\) to side \(AC,\) which is parallel to side \(BC.\).
From the point \(C,\) draw a line \(CM\) parallel to \(AB.\)
Extends the \(DE,\) line that intersects the \(CM\) line at \(F.\).
Given \(DE∥BC,\) this implies \(DF∥BC\) and \(BD∥CF\) (by construction)
Thus the quadrilateral \(BDFC\) is a parallelogram.
\(BD=CF\) (By C.P.C.T)
From i, \(AD=CF\)
By construction, the lines \(BD\) and \(CF\) are parallel, and the line \(DF\) is the transversal.
The pair of alternating angles formed by the parallel lines with the transversal are equal.
So \(\Angle ADE = \Angle CFE\)
Well, in the triangles \(ADE\) and \(ECF,\)
\(AD=CF\) (proved above)
\(\angle ADE = \angle CFE\) (alternative interior angles)
\(\angle AED = \angle CEF\) (vertically opposite angles)
According to the A.S.A congruence rule, two triangles \(ADE\) and \(CEF,\) are congruent.
\(△ADE≅△CEF\)
The Pages \(AE=EC\) (By C.P.C.T)
Therefore, point \(E\) is the midpoint of side \(AC.\)
Solved examples
Q.1. \(l, m,\) and \(n\) are three parallel lines. \(p\) and \(q\) are the transversals intersecting parallel lines at \(A, B, C, D, E,\) and \(F,\) as shown in the figure. If \(AB:BC= 1:1,\) find the ratio of \(DE:EF.\)
Answer: Given \(AB:BC=1:1\)
Join \(AF,\) that intersects the line m at the point \(G.\).
In \(△ACF,\)
\(AB:BC=1:1,\) which means that \(B\) is the center of \(AC.\).
And, \(BG∥CF,\) as given, the lines \(m\) and \(n\) are parallel.
By reversing the midpoint theorem, the line drawn from the midpoint parallel to the other side bisects the third side of the triangle.
Therefore \(G\) is the midpoint of the side \(AF.\)
So \(AG=GF.\)
Well, in \(△AFD,\)
\(AG=GF\) (proved above) and \(GE∥AD\) since the lines l and m are parallel.
Therefore, after inverting the midpoint theorem \(E\), the midpoint of the side \(DF.\)
So \(DE=EF\)
Therefore the ratio \(DE:EF=1:1.\)
F.2. Find the perimeter of the triangle formed by joining \(L, M,\) and \(N\). Where \(L, M\) and \(N\) are the midpoints of the side \(PQ, QR,\) and \(PR\), respectively, of the triangle \(PQR\) and \(PQ = 8\; {\ rm{cm}},QR = 9\;{\rm{cm}}\) and \(PR = 6\,{\rm{cm}}{\rm{.}}\)
Answer: Since \(L, M\) and \(N\) are the midpoints of the sides \(PQ, QR,\) and \(PR\) of the triangle \(PQR.\).
\(PQ = 8\;{\rm{cm}},QR = 9\;{\rm{cm}}\) and \(PR = 6\,{\rm{cm}}{\rm{ .}}\)
According to the midpoint theorem, the line joining the midpoints of any two sides is parallel to the third side and equal to half the third side.
In the triangle \(PQR,\)
\(LM∥PQ\) and \(LM = \frac{1}{2}PQ = \frac{1}{2}(8) = 4\;{\rm{cm}}\)
\(LN∥PR\) and \(LN = \frac{1}{2}PR = \frac{1}{2}(6) = 3\;{\rm{cm}}\)
\(MN∥QR\) and \(MN = \frac{1}{2}QR = \frac{1}{2}(9) = 4.5\;{\rm{cm}}\)
F.3. The values given in the figure below are given in centimeters. Side length \(AX = 9\,{\rm{cm}}{\rm{.}}\) Determine the side length \(AO\) using the midpoint theorem.
Answer: From the given number,
\(AD = DB = 4\;{\rm{cm}}}\) and \(AE = EC = 6\;{\rm{cm}}}\)
Therefore, the points \(D\) and \(E\) are the midpoints of the sides \(AB\) and \(AC.\)
According to the midpoint theorem, the line connecting the points \(D\) and \(E\) is parallel to the side \(BC\) and equal to half of \(BC.\)
\(DE∥BC,\) which implies \(DO∥BX.\).
Now in the triangle \(ABX,\) the line \(DO\) is drawn from the center \(D\) of the side \(AB\) and parallel to the side \(BX.\)
By inverting the midpoint theorem, we know that the line drawn from the midpoint of one side is parallel to the other side, bisecting the third side of the triangle.
Thus the line \(DO\) bisects the third line \(AX.\)
\(AO = \frac{1}{2}AX = \frac{1}{2}(9\;{\rm{cm}})) = 4.5\;{\rm{cm}}\)
Therefore the value of \(AO\) is \(4.5\;{\rm{cm}}.\)
F.4. In a triangle \(ABC, X\) and \(Y\) are the midpoints of the sides \(AC\) and \(BC.\) Find the value of \(XY.\)
Answer: Given, \(AB = 10,AC = 8\) and \({\rm{BC = 6\, units}}{\rm{.}}\)
Also, the points \(X\) and \(Y\) are the midpoints of the sides \(AC\) and \(BC.\)
According to the midpoint theorem, the line joining the midpoints of any two sides is parallel to the third side and equal to half of it.
So the line \(XY\) is parallel to \(AB\) and equal to half of \(AB.\)
\(XY = \frac{1}{2}AB = \frac{1}{2}(10) = 5\,{\rm{units}}\)
F.5. Find the value of the hypotenuse of the triangle \(ABC,\) in the \(AC = 8\,{\rm{units}}\) and \({\rm{BC = 6 units}}\) and the length of the Line segment connecting midpoints \(X\) and \(Y\) is \(5\) units using the midpoint theorem.
Answer: Given, \(AB = 10,AC = 8\) and \({\rm{BC = 6 \,units}}{\rm{.}}\)
The length of the line segment connecting the midpoints \(X\) and \(Y\) of the sides \(AC\) and \(BC\) are \(5\) units.
According to the midpoint theorem, the line joining the midpoints of any two sides is parallel to the third side and equal to half of it.
So the line \(XY\) is parallel to \(AB\) and equal to half of \(AB.\)
\(XY = \frac{1}{2}AB\) units
\({\rm{AB = 2 XY = 2(5) = 10 \.units}}\)
summary
In this article, we have examined the statement and proof of the midpoint theorem and the statement and proof of the converse of the midpoint theorem. The midpoint theorem states that in a triangle, if a line connecting midpoints of any two sides of the triangle is parallel to the third side of the triangle and its value is half the measure of the third side of the triangle.
The converse of the missing point theorem states that the line drawn through the midpoint of any side of the triangle parallel to the other side of the triangle bisects the third side of the triangle. This article contains the solved examples of midpoint theorem and inverse midpoint theorem, which will help us to understand it easily.
Learn more about the midpoint formula
frequently asked Questions
Q.1. What is the midpoint?
Answer: The midpoint is the point on the line segment that is equidistant from either end of the line segment.
F.2. Define midpoint theorem?
Answer: In a triangle, when a line joining the midpoints of any two sides of the triangle is parallel to the third side of the triangle and its value is equal to half the measure of the third side of the triangle.
F.3. What is the inverse of the midpoint theorem?
Answer: The line drawn through the midpoint of any side of the triangle parallel to the other side of the triangle bisects the third side of the triangle.
F.4. Is the midpoint theorem an application of Thales’ theorem?
Answer: Yes. The midpoint theorem is the application of Thales’ theorem.
F.5. If the inverse of the midpoint theorem is the application of the inverse of B.P.T.
Answer: Yes. The inverse of the midpoint theorem is the application of the inverse of B.P.T.
Whats the difference between a bisector and a midpoint?
Midpoints are points exactly in the middle of a segment: they’re equidistant to either end. Segment bisectors are lines that cut a segment right in half, which means they go through the midpoint of the segment.
Prove that the lines parallel to the angle bisectors through the midpoints of sides of a triangle are concurrent at the Spieker center
Midpoints are points exactly in the middle of a segment: they are equidistant from either end. Segment bisectors are lines that cut a segment directly in half, meaning they pass through the midpoint of the segment.
How do you prove a line is a bisector?
A line that splits another line segment (or an angle) into two equal parts is called a “bisector.” If the intersection between the two line segment is at a right angle, then the two lines are perpendicular, and the bisector is called a “perpendicular bisector”.
Prove that the lines parallel to the angle bisectors through the midpoints of sides of a triangle are concurrent at the Spieker center
A line that divides another line segment (or angle) into two equal parts is called a “bisector”. If the intersection between the two line segments is at right angles, the two lines are perpendicular and the bisector is called the “perpendicular bisector”.
The bisector theorem states that a point on the bisector of a line segment is equidistant from both edges of the line segment.
problem
Point C lies on the perpendicular bisector of line AB. Prove that |CA|=|CB|
strategy
We need to prove that two line segments are equal, so let’s draw them:
And now it should be clear that the way to go here is to use congruent triangles. We have a common side (CD). We have also given two segments equal (AD=DB) since AD is the bisector. And we have an equal angle, which is a right angle, between them.
prove
(1) CD=CD //Common side, reflexive equality property
(2) AD=DB //Given CD is an angle bisector
(3) m∠ADC= m∠BDC =90° //Given that CD is perpendicular to AB
(4) ∠ADC≅ ∠BDC //aspect-angle-aspect postulate
(5) CA=CB //corresponding sides of congruent triangles (CPCTC)
We have thus proved the angle bisector theorem. Note that this is an inverse of one of the properties of an isosceles triangle – that the median to the base (which is a bisector) is perpendicular to the base. Here we prove the opposite – if we have a perpendicular median, the triangle is isosceles.
What do you call the line that bisects a segment on its midpoint?
A line, segment, or ray that passes through a midpoint of another segment is called a segment bisector. A bisector cuts a line segment into two congruent parts. A segment bisector is called a perpendicular bisector when the bisector intersects the segment at a right angle.
Prove that the lines parallel to the angle bisectors through the midpoints of sides of a triangle are concurrent at the Spieker center
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Is midpoint line perpendicular?
Key Points
We know that the perpendicular bisector of a line segment is the unique line perpendicular to the segment passing through its midpoint.
Prove that the lines parallel to the angle bisectors through the midpoints of sides of a triangle are concurrent at the Spieker center
Our first goal is to learn how to calculate the coordinates of the midpoint of a line segment connecting two points. Suppose we get two points 𝑃 ( 𝑥 , 𝑦 ) and 𝑃 ( 𝑥 , 𝑦 ) . The midpoint of the segment 𝑃 𝑃 is the point 𝑀, which is on 𝑃 𝑃 exactly midway between 𝑃 and 𝑃 . This means that the 𝑥 coordinate 𝑥 of 𝑀 is halfway between 𝑥 and 𝑥 and can therefore be calculated by averaging the two points, giving us 𝑥 = 𝑥 + 𝑥 2 . The same is true for the 𝑦 coordinate 𝑦 of 𝑀 . This leads us to the following formula.
Formula: The coordinates of a midpoint Suppose 𝑃 ( 𝑥 , 𝑦 ) and 𝑃 ( 𝑥 , 𝑦 ) are points connected by a line segment 𝑃 𝑃 . Then the coordinates of the midpoint of the line segment 𝑃 𝑃 are given by 𝑀 ( 𝑥 , 𝑦 ) = 𝑥 + 𝑥 2 , 𝑦 + 𝑦 2 .
Let’s practice finding the coordinates of midpoints.
Example 1: Finding the midpoint of a line segment given endpoints Given 𝐴 ( 4 , 8 ) and 𝐵 ( 6 , 6 ) , what are the coordinates of the midpoint of 𝐴 𝐵 ? Answer Recall that the midpoint 𝑀 of a line segment is the midpoint between the endpoints, which we can find by averaging the 𝑥 and 𝑦 coordinates of 𝐴 and 𝐵, respectively. Therefore, we apply the formula: 𝑀 ( 𝑥 , 𝑦 ) = 𝑥 + 𝑥 2 , 𝑦 + 𝑦 2 = 4 + 6 2 , 8 + 6 2 = ( 5 , 7 ) . Therefore the coordinates of the center of 𝐴 𝐵 are (5, 7).
We can also use the formula for the coordinates of a midpoint to calculate one of the endpoints of a line segment, given its other endpoint and the coordinates of the midpoint.
Example 2: Finding an endpoint of a line segment given the midpoint and the other endpoint The origin is the midpoint of the straight segment 𝐴 𝐵 . Find the coordinates of point 𝐵 when the coordinates of point 𝐴 are (−6, 4). Answer Here we got one endpoint of a line segment and the midpoint 𝑀 ( 0 , 0 ) and were asked to find the other endpoint. We can do this by using the midpoint formula in reverse: 𝑀 ( 𝑥 , 𝑦 ) = ( 0 , 0 ) = 𝑥 + 𝑥 2 , 𝑦 + 𝑦 2 = − 6 + 𝑥 2 , 4 + 𝑦 2 . This gives us two equations: − 6 + 𝑥 2 = 0 − 6 + 𝑥 = 0 𝑥 = 6 and 4 + 𝑦 2 = 0 4 + 𝑦 = 0 𝑦 = − 4 . We conclude that the coordinates of 𝐵 are ( 6 , − 4 ).
One application for calculating the centers of line segments is to calculate the coordinates of the centers of circles considering their diameters for the simple reason that the center of a circle is the center of each of its diameters.
In the next example we see an example of finding the center of a circle using this method.
Example 3: Finding the center of a circle given endpoints of a diameter The points 𝐴 ( 4 , 1 ) and 𝐵 ( − 4 , − 5 ) define the diameter 𝐴 𝐵 of a circle with center 𝑀 . Find the coordinates of 𝑀 and the circumference of the circle and round your answer to the nearest tenth. Answer The center 𝑀 of the circle is the center of its diameter 𝐴 𝐵 . Remember that the midpoint 𝑀 of a line segment (e.g. a diameter) can be found by averaging the 𝑥 and 𝑦 coordinates of the endpoints 𝐴 and 𝐵 as follows: 𝑀 ( 𝑥 , 𝑦 ) = 𝑥 + 𝑥 2 , 𝑦 + 𝑦 2 = 4 − 4 2 , 1 − 5 2 = ( 0 , − 2 ) . The circumference of a circle is given by the formula 𝐶 = 2 𝜋 𝑟, where 𝑟 is the length of its radius. The length of the radius is the distance from the center of the circle to any point on its radius, for example point 𝐴(4, 1). We can calculate this length using the formula for the distance between two points 𝑀 and 𝐴: 𝑟 = ( 𝑥 − 𝑥 ) + ( 𝑦 − 𝑦 ) 𝑟 = ( 0 − 4 ) + ( − 2 − 1 ) = 1 6 + 9 = 2 5 . By taking the square roots we find that 𝑟 = 5 and therefore the perimeter is 2 𝜋 𝑟 = 1 0 𝜋 = 3 1. 4 to the nearest tenth. In summary, the coordinates of the center are ( 0 , − 2 ) and the perimeter is 31.4 to the nearest tenth.
We now turn to the second main subject of this explainer, the computation of the equation of the perpendicular bisector of a given line.
Definition: Normal Bisector For a line segment 𝐴 𝐵, the normal bisector of 𝐴 𝐵 is the unique line perpendicular to 𝐴 𝐵 that passes through the midpoint of 𝐴 𝐵.
Recall that for any line 𝐿 with slope 𝑚, the slope of each line perpendicular to it is the negative reciprocal of 𝑚, i.e. − 1 𝑚 . We can use this fact and our understanding of midpoints of line segments to write down the equation of the perpendicular bisector of any line segment.
How to: Calculate the Equation of the Bisector of a Line Suppose we have a line segment 𝑃 𝑃 with endpoints 𝑃 ( 𝑥 , 𝑦 ) and 𝑃 ( 𝑥 , 𝑦 ) and we want to find the equation of its bisector . First we calculate the slope of the line segment. To do this, we recall the definition of slope: slope of change 𝑃 𝑃 = 𝑦 𝑥 = 𝑦 − 𝑦 𝑥 − 𝑥 . Next we calculate the slope of the perpendicular bisector as the negative reciprocal of the slope of the segment: Next we find the coordinates of the center of 𝑃 𝑃 by applying the formula to the endpoints: 𝑀 ( 𝑥 , 𝑦 ) = 𝑥 + 𝑥 2 , 𝑦 + 𝑦 2 . We can now plug these coordinates and the slope into the point-slope form of the line equation: ( 𝑦 − 𝑦 ) = 𝑚 ( 𝑥 − 𝑥 ) 𝑦 − 𝑦 + 𝑦 2 = − 𝑥 − 𝑥 𝑦 − 𝑦 𝑥 − 𝑥 + 𝑥 2 .
This gives us an equation for the perpendicular bisector.
Let’s try to apply this algorithm.
Example 4: Find the perpendicular bisector of a line segment connecting two points Find the equation of the perpendicular bisector of the line segment connecting points 𝐴 ( 1 , 3 ) and 𝐵 ( 7 , 1 1 ). Give your answer in the form 𝑦 = 𝑚 𝑥 + 𝑐 . Answer To find the equation of the perpendicular bisector, we must first find its slope, which is the negative reciprocal of the slope of the line segment connecting 𝐴 and 𝐵. This is given by slope of 𝐴 𝐵 = 𝑦 − 𝑦 𝑥 − 𝑥 = 1 1 − 3 7 − 1 = 8 6 = 4 3 . Now we can find the negative reciprocal by flipping the fraction and taking the negative; this gives us the following: s l o p e o f p e r p e n d i c u l a r b i s e c t o r = − 3 4 . Next we need the coordinates of a point on the perpendicular bisector. Since the perpendicular bisector (by definition) goes through the midpoint of the segment 𝐴 𝐵, we can use the formula for the coordinates of the midpoint: 𝑀 ( 𝑥 , 𝑦 ) = 𝑥 + 𝑥 2 , 𝑦 + 𝑦 2 = 1 + 7 2 , 3 + 1 1 2 = ( 4 , 7 ) . Plugging these coordinates and our slope − 3 4 into the point-slope form of the straight line equation, 𝑦 − 𝑦 = 𝑚 ( 𝑥 − 𝑥 ) 𝑦 − 7 = − 3 4 ( 𝑥 − 4 ) , and rearranging to the form 𝑦 = 𝑚 𝑥 + 𝑐 , we have 𝑦 = − 3 4 𝑥 + 1 0 .
For our final example, we’ll use our understanding of midpoints and perpendicular bisectors to calculate some unknown values.
Example 5: Determining the unknown variables that describe a perpendicular bisector of a segment A segment 𝐴 𝐵 connects the points 𝐴 ( − 6 , − 6 ) and 𝐵 ( 0 , 𝑝 ) . The perpendicular bisector of 𝐴 𝐵 has the equation 𝑦 = − 3 𝑥 + 𝑐 . Find the values of 𝑝 and 𝑐 . Answer Since the bisector has slope − 3, we know that line 𝐴 𝐵 has slope 1 3 (the negative reciprocal of − 3 ). We can calculate the 𝑦 coordinate of point 𝐵 (ie 𝑝 ) using the definition of slope 𝑚: 𝑚 = 𝑦 − 𝑦 𝑥 − 𝑥 1 3 = 𝑝 − ( − 6 ) 0 − ( − 6 ) = 𝑝 + 6 6 2 = 𝑝 + 6 𝑝 = − 4 . We calculate the value of 𝑐 in the equation 𝑦 = − 3 𝑥 + 𝑐 of the perpendicular bisector using the coordinates of the center 𝑀 of 𝐴 𝐵 (which by definition is a point lying on the perpendicular bisector). We have the formula 𝑀 ( 𝑥 , 𝑦 ) = 𝑥 + 𝑥 2 , 𝑦 + 𝑦 2 = − 6 + 0 2 , − 6 − 4 2 = ( − 3 , − 5 ) . We can now substitute 𝑥 = − 3 and 𝑦 = − 5 into the perpendicular bisector equation and rearrange to find 𝑐: 𝑦 = − 3 𝑥 + 𝑐 − 5 = − 3 × − 3 + 𝑐 𝑐 = − 5 − 9 = − 1 4 . Our solution for the example is 𝑝 = − 4 , 𝑐 = − 1 4 .
Finally, let’s recap some key concepts from this explainer.
Angle Bisector Theorem – Midpoints \u0026 Line Segments
See some more details on the topic geometry proofs with midpoints and angle bisectors here:
Geometry Proofs with Midpoints & Angle Bisectors
Beginning Proofs – Geometry Proofs with Mpoints & Angle Bisectors. 1. Complete the following proof: Given: L is the mpoint of KM.
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1. Use the midpoint and angle bisector theorems to justify …
Today’s objectives: 1. Use the mpoint and angle bisector theorems to justify geometric statements. 2. Use previously learned postulates, theorems, and.
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Perpendicular Bisector Theorem – Proofs, Solved Examples
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Postulates, Theorems, and Proofs Goals – Madeira City Schools
Subject: Honors Geometry. Date: «date». 1. A Definition of Mpoint. B Definition of Angle Bisector. C Angle Addition Postulate.
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Mid Point Theorem- Statement, Proof, Formula, and Example
midpoint theorem
Geometry is one of the most important and essential branches of mathematics. This field deals with geometric problems and figures based on their properties. One of the most important theorems in geometry dealing with the properties of triangles is called the midpoint theorem.
Table of Contents:
The theory of the midpoint theorem is used in coordinate geometry and states that the midpoint of the line segment is an intersection of the endpoints. Both the “x” and “y” coordinates must be known to solve an equation using this theorem. The midpoint theorem is also useful in calculus and algebra.
MidPoint theorem statement
The midpoint theorem states, “The line segment in a triangle joining the midpoint of two sides of the triangle shall be parallel to its third side and shall also be half the length of the third side.”
Midpoint theorem proof
When the line segment is adjacent to the midpoints of one of the sides of a triangle, the line segment is said to be parallel to all remaining sides and spans about half the remaining sides.
Consider triangle ABC as shown in the figure above,
Let E and D be the midpoints of sides AC and AB. Then the line DE is said to be parallel to side BC, while side DE is half of side BC; i.e.
\(\begin{array}{l}DE \parallel BC\end{array} \)
DE = (1/2 * BC).
Now look at the following figure,
Construction – Extend line segment DE and produce to F such that EF = DE.
In the triangle ADE and CFE,
EC = AE —– (given)
∠CEF = ∠AED (perpendicularly opposed angles)
∠DAE = ∠ECF (alternating angle)
According to the ASA congruence criterion
△ CFE ≅ △ ADE
Because of this,
∠CFE = ∠ADE {through c.p.c.t.}
∠FCE= ∠DAE {through c.p.c.t.}
and CF = AD {through c.p.c.t.}
∠CFE and ∠ADE are the alternating interior angles.
Assume CF and AB as two lines intersected by the transverse DF.
Similarly, ∠FCE and ∠DAE are the alternating interior angles.
Suppose CF and AB are the two lines intersected by the transverse AC.
So CF ∥ AB
So CF ∥ BD
and CF = BD {since BD = AD, CF = AD is proved}
Thus BDFC forms a parallelogram.
Through the properties of a parallelogram we can write
BC ∥ DF
and BC = DF
BC ∥ DE
and DE = (1/2 * BC).
This proves the midpoint theorem.
MidPoint theorem formula
In coordinate geometry, the midpoint set refers to the midpoint of the line segment. It defines the coordinate points of the midpoint of the line segment and can be found by averaging the coordinates of the specified endpoints. The midpoint formula is used to find the midpoint between the two given points.
If P 1 (x 1 ,y 1 ) and P 2 (x 2 ,y 2 ) are the coordinates of two given endpoints, then the midpoint formula is given as:
Midpoint = [(x 1 + x 2 )/2, (y 1 + y 2 )/2]
The inverse of the midpoint theorem
The converse of the midpoint theorem states that “when a line is drawn through the midpoint of one side of a triangle and is parallel to the other side, it bisects the third side”.
Example of the midpoint theorem
The example is given below to understand the midpoint theorem.
Example:
In triangle ABC, the midpoints of BC, CA, AB are D, E, and F, respectively. Find the value of EF when the value of BC = 14 cm
Solution:
Given: BC = 14 cm
If F is the midpoint of AB and E is the midpoint of AC, then using the midpoint theorem:
EF = 1/2 (BC)
replacing the value of BC,
EF = (1/2) × 14
EF = 7 cm
Therefore the value of EF = 7 cm.
The midpoint theorem can also be proved using triangles. Suppose two lines are drawn parallel to the x and y axes, starting at endpoints and connected through the midpoint, then the segment passes through the angle between them, resulting in two similar triangles. This relationship of these triangles forms the midpoint theorem.
1.4: Midpoints and Segment Bisector
Use midpoints and bisectors to find the midpoint between two coordinates.
When two segments are congruent, we indicate that they are congruent or equal in length with segment markers as shown below:
Figure \(\PageIndex{1}\)
A midpoint is a point on a line segment that divides it into two congruent segments.
Figure \(\PageIndex{2}\)
Since \(AB=BC\), \(B\) is the midpoint of \(\overline{AC}\). Each line segment has exactly one midpoint.
When plotting points in the coordinate plane, we can use a formula to find the midpoint between them.
Here are two points, \((-5, 6)\) and \((3, 2).\)
Figure \(\PageIndex{3}\)
The midpoint should be halfway between the points on the segment connecting them. If you just look, it looks like the midpoint is \((-1, 4).\)
Midpoint formula: For two points, (\(x_1,y_1\)) and (\(x_2,y_2\)), the midpoint is \(\left (\dfrac{x_1+x_2}{2} , \dfrac{y_1 + y_2}{2}\right) \).
Let’s use the formula to ensure that \((-1, 4)\) is the midpoint between \((-5, 6)\) and \((3, 2)\).
\(\left (\dfrac{-5+3}{2} ,\dfrac{6+2}{2}\right) =(−22.82)=(−1.4) \)
A bisector cuts a line segment into two congruent parts and passes through the midpoint. A bisector is a segment bisector that intersects the segment at right angles.
\(\overline{AB} \cong \overline{BC}\)
\(\overline{AC} \perp \overleftrightarrow{DE}\)
Figure \(\PageIndex{4}\)
What if you were given the coordinates of two points and you wanted to find the point right in the middle? How would you find the coordinates of this third point?
Example \(\PageIndex{1}\) Write all the same segment statements. Figure \(\PageIndex{5}\) Solution \(AD=DE\) \(FD=DB=DC\)
Example \(\PageIndex{2}\) Is \(M\) a midpoint of \(\overline{AB}\)? Figure \(\PageIndex{6}\) Solution No, it’s not \(MB=16\) and \(AM=34−16=18\). \(AM\) must equal \(MB\) so that M is the midpoint of \(\overline{AB}\).
Example \(\PageIndex{3}\) Find the midpoint between \((9, -2)\) and \((-5, 14)\). Solution Plug the points into the formula. \(\left(\dfrac{9+(−5)}{2}\ ,\dfrac{−2+14}{2}\right)=\left(\dfrac{4}{2},\dfrac{ 12}{2}\right)=(2,6)\)
Example \(\PageIndex{4}\) Which line is the perpendicular bisector of \(\overline{MN}\)? Figure \(\PageIndex{7}\) Solution The perpendicular bisector must bisect \(\overline{MN}\) and be perpendicular to it. Only \(\overleftrightarrow{OQ}\) fits this description. \(\overleftrightarrow{SR}\) bisects an angle, but is not perpendicular.
Example \(\PageIndex{5}\) Find \(x\) and \(y\). Figure \(\PageIndex{8}\) Solution The straight line shown is the perpendicular bisector. So \(3x−6=21\) \(3x=27\) \(x=9\) And \((4y−2)=90\) \(4y=92\) \(y=23 \)
Verification Copy the following figure and label it with the following information: \(\overline{AB} \cong \overline{AD}\) \(\overline{CD} \cong \overline{BC}\) figure \( \PageIndex {9}\) For 2-4, use the following image to answer the questions. Figure \(\PageIndex{10}\) \(P\) is the center of which two segments? How does \(\overline{VS}\) relate to \(\overline{QT}\)? How does \(\overline{QT}\) relate to \(\overline{VS}\)? For Exercise 5, use algebra to find the value of \(x\). Figure \(\PageIndex{11}\) For questions 6-10, find the midpoint between each pair of dots. (-2, -3) and (8, -7) (9, -1) and (-6, -11) (-4, 10) and (14, 0) (0, -5) and (-9 , 9) (-3, -5) and (2, 1) Given the midpoint (\(M\)) and one of the endpoints of \(\overline{AB}\), find the other endpoint. \(A(−1,2)\) and \(M(3,6)\) \(B(−10,−7)\) and \(M(−2,1)\)
Review (Responses) To view the review responses, open this PDF and locate Section 1.4.
Vocabulary Term Definition Midpoint The midpoint of a line segment is the point on the line segment that divides the segment into two congruent parts. Bisector A segment bisector that intersects the segment at right angles. Segment bisector A segment bisector is a line (or part of a line) that passes through the midpoint. Segment markers When two segments are congruent, we indicate that they are congruent with segment markers. Midpoint formula The midpoint formula states that for the endpoints \((x_1, y_1)\) and \((x_2, y_2)\), the midpoint \( \left (\dfrac{x_1+x_2}{2}, \dfrac {y_1 +y_2}{2}\right)\).
Prove that the lines parallel to the angle bisectors through the midpoints of sides of a triangle are concurrent at the Spieker center
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The parallelism is evident from the medial triangle property as stated in cosmo5’s answer. Here is a wiki link you can refer to learn more about it (wiki).
Here is a more elementary proof that simply uses i) the midpoint theorem, ii) the alternating interior angles theorem, which states that when two parallel lines are intersected by a traversal, the alternating interior angles are congruent, and iii) the theorem about the simultaneity of the bisecting lines.
$D, E$ and $F$ are midpoints of the sides of $\triangle ABC$. The line through $AM$ is the bisector of $\angle A$. The line through $EN$ is parallel to $AM$ and through the midpoint $E$ of side $BC$.
Using the midpoint theorem, $ADEF$ is a parallelogram and $\angle DEF = \angle A$.
Using the parallel line property, $\angle FEN = \angle FMA = \angle MAD = \frac{\angle A}{2}$. So the line through $EN$ is the bisector of $\angle DEF$ in $\triangle DEF$. By the same logic, parallel lines through midpoints to two other bisectors of $\triangle ABC$ must be bisectors of $\angle DFE$ and $\angle EDF$.
Hence the proof is complete by simultaneity of the bisectors of a triangle.
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