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Indirect Proof in Geometry: Definition & ExamplesNext Lesson
Proving the Triangle Midsegments Theorem
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Triangle Midsegment Theorem
Suppose there are four roads in your neighborhood, such that three of the roads form a triangle, and the fourth road (Smith Street) connects two of the other roads at their midpoints.
Did you know there is an actual theorem about Smith Street? The Triangle Midsegment Theorem states that the line segment connecting the midpoints of any two sides of a triangle will satisfy the following properties:
 The line segment will be parallel to the third side.
 The length of the line segment will be onehalf the length of the third side.
In terms of the roads, this tells us that Smith Street is parallel to Bradford Road, and that Smith Street is onehalf the distance of Bradford Road. Neat! We can use the theorem to determine that Smith Street has length 2 km, since 1/2 × 4 = 2.
The proof of this theorem involves proving both of the properties using a series of logical deductions along with several other theorems and properties. We definitely want to see this proof, so let’s first familiarize ourselves with some facts and properties that we’ll need to use to carry out the proof.
Facts for the Proof
Get ready for a lot of information at once. However, it will be not only useful, but essential to have these to look at when we do the proof. If you’re unfamiliar with why these facts are true, don’t worry. They’re just here for reference. We’re more interested in the facts themselves.
In a triangle ABC, if we connect the midpoints, D and E, of any two sides, say AB and BC, as shown in the image, then the following facts are true:
 AD = BD and CE = BE, because D and E are the midpoints of AB and BC, respectively.
 Side Angle Side Triangle Similarity Theorem: Triangles are similar if the ratio between two corresponding sides is equal to the ratio between another two corresponding sides, and the angles between those corresponding sides have the same measure.
 Similar triangles have corresponding angles of equal measure, and corresponding sides are proportional, or create the same ratio.
 Converse of the Corresponding Angles Postulate: If we draw a line, called a transversal, through any two lines, and it creates two corresponding angles that have equal measure, then the two lines are parallel.
Okay, I told you it was a lot of information all at once, but we’re going to be very glad that we have it handy when we do this proof. Speaking of which, we’re all set! Let’s do this!
Proof of the Triangle Midsegment Theorem
Given a triangle ABC, let’s draw a line segment connecting the midpoints of two of the sides, say AB and BC.
To prove the Triangle Midsegment Theorem, we need to show two things:
 DE is onehalf of AC. That is, DE = (1/2)AC.
 DE is parallel to AC.
We’ll start with part one. Because D is the midpoint of AB, we have that AD = DB. Therefore,
 AB = AD + DB = DB + DB = 2DB
Similarly, since E is the midpoint of BC, we have that BE = EC, and
 BC = BE + EC = BE + BE = 2BE
Therefore, we have the following:
 DB / AB = BE / BC = 1/2
Consider the ΔABC and ΔDBE. We just showed that the ratio between corresponding sides DB and AB and between corresponding sides BE and BC are both equal to 1/2. Also, notice that these two triangles share the angle B, so we know the measure of the angle between these two sets of corresponding sides have equal measure. Therefore, by the Side Angle Side Theorem for Similar Triangles, we have that ΔABC is similar to ΔDBE.
Since ΔABC is similar to ΔDBE, we have that all of the corresponding sides of these two triangles are proportional and create the ratio 1/2.
 DB / AB = BE / BC = DE / AC = 1/2
Now look at the ratio DE / AC = 1/2, specifically. If we multiply both sides of this equation by AC, we get the following:
 DE = (1/2)AC
This is exactly what we wanted to prove for part one! All we have left is part two. That is, we need to prove that DE is parallel to AC.
Because ΔABC is similar to ΔDBE, we have that the two corresponding angles, ∠BDE and ∠BAC have the same measure. Notice these are also two corresponding angles if we were to draw the line, or transversal, AB through the two lines AC and DE. Therefore, by the Converse of the Corresponding Angles Postulate, we have that DE must be parallel to AC, and this proves part two!
Tada! We’ve proven both parts of the Triangle Midsegment Theorem, so we’ve proven the theorem! That wasn’t so hard, was it?
Lesson Summary
The Triangle Midsegment Theorem states that, if we connect the midpoints of any two sides of a triangle with a line segment, then that line segment satisfies the following two properties:
 The line segment will be parallel to the third side.
 The length of the line segment will be onehalf the length of the third side.
The proof of this theorem is simply a matter of using properties of similar triangles and corresponding angles to logically deduce that both of the properties in the theorem are true.
This theorem is extremely useful in many realworld applications. The next time we need to use this theorem, we not only know the theorem and how it works: thanks to the proof of the theorem, we also know why it works.
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Proving the Triangle Midsegments Theorem
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 Math Geometry Triangle Triangle Mid Segment Theorem
Top
Triangle Mid Segment Theorem
Triangle is a most important two dimensional polygon in mathematics. In geometry, we deal with various geometric figures like rectangle, square, cube, parallelogram etc. Among all triangles are polygons with three vertices and three edges. Many practical problems are based on triangle theorems such as SideSplitter Theorem, Angle Bisector Theorem and Area Ratios of triangle.
Triangle mid segment is one of the basic theorems on triangles which explains a very important property of triangles. Before working on midsegment theorem first we should know about “midsegment”, A midsegment of a triangle connects the midpoints of two sides of a triangle. In this section will study about triangle mid segment theorem in detail.
Related Calculators  
Mid Point Calculator  Mid Quartile Calculator 
Bayes Theorem Calculator  Binomial Theorem Calculator 
Proof
Back to Top
Statement: A segment that joins the mid points of any two sides is measure half of the length of third side and also parallel to it.
That is, ST = $\frac12$ QR and ST ll QR.
Proof:
Since S and T are mid points of triangle.
SP = QS and TP = RT
Also PR = 2PT and PQ = 2PS
Consider two triangles, $\triangle$PST and $\triangle$PQR,
$\fracPSPQ$ = $\fracPTPR$ = $\frac12$ (angle P is common)
Which shows that both the triangles are similar.
or $\triangle$PST $\sim$ $\triangle PQR$
Since ST and QR are corresponding parts of the similar triangles, which implies
$ \fracPSPQ$ = $\fracPTPR$ = $\fracSTQR$ = $\frac12$
or $\fracSTQR$ = $\frac12$
or ST = $\frac12$ (QR)
Also $\angle$ PTS = $\angle$ PRQ (corresponding angles of similar triangles)
Transversal line PR cuts line segments ST and QR (By transversal property)
Using above results, we obtained
ST ll QR
Hence Proved.
Converse
Back to Top
Statement: The straight line passing through the midpoint of one leg of a triangle bisect another leg at its midpoint.
Say S is the midpoint of PQ
To prove: T is midpoint of PR
Extends line passing through S till V and join RV in such a way
QS  RV and SV  QR
Proof:
In parallelogram QSVR,
QS = RV
Also PS = QS (S is mid point of PQ)
=> RV = PS …(1)
In $\triangle$ PST and $\triangle$ RTV:
$\angle P$ = $\angleR$ (RV QS) …(2)
$\anglePTS$ = $\angleRTV$ (vertical opposite angles) …(3)
From (1), (2) and (3)
$\triangle$ PST $\cong$ $\triangle$ RTV
=> PT = TR
=> T is midpoint of PR.
Examples
Back to Top
Below are some examples on mid segment theorem:
Example 1: If the vertices of any triangle are (3, 5), (1, 3) and (5, 2). Show that mid segment is half the length to the side opposite to it.
Solution: Consider a triangle ABC with vertex coordinates A(3, 5), B(1, 3) and C (5, 2).
Step 1: Find the midpoints of two sides
Consider E and F are midpoints of AB and AC respectively
Midpoint of AB = $(\frac3+12, \frac5+32)$ = (2, 4) or E(2, 4)
Midpoint of AC = $(\frac3+52, \frac522)$ = (4, 1.5) or F(4, 1.5)
Step 2: Measure of EF and BC (using distance formula)
EF = $\sqrt(42)^2 + (1.5 – 4)^2$ = $\sqrt4 + 6.25$ = $\sqrt10.25$ = 3.202
BC = $\sqrt(51)^2 + (2 – 3)^2$ = $\sqrt16 + 25$ = $\sqrt41$ = 6.403
Step 3:
EF = $\frac12$ (BC) = $\frac12$ $\times$ 6.403 = 3.202 (True, from step 2)
Example 2: Find the value of m. ST is parallel to RQ.
Solution: Since mid segment of a triangle is half the length of its parallel side.
That is, ST = $\frac12$(RQ)
ST = m and RQ = 30 (given)
Substitute the number into above equation
=> m = $\frac12$ $\times$ 30
or m = 15
The value of m is 15 units.
Example 3: In $\triangle$ XYZ, if R is the midpoint of XY and S is the midpoint of XZ. Also RS  YZ. Find the
lengths of RS and YZ if RS = x – 3 and YZ = 4x – 16.
Solution: Since RS  YZ
RS = x – 3 and YZ = 4x – 16 (given)
Using mid segment theorem, we get
x – 3 = $\frac12$ (4x – 16)
x – 3 = 2x – 8
x = 5
Now RS = x – 3 = 5 – 3 = 2 and
YZ = 4x – 16 = 20 – 16 = 4
=> RS = 2 and YZ = 4
Therefore, length of RS is 2 units and length of YZ is 4 units.
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 Home
 How it works
 About Us
 Algebra Booster !
 Math Geometry Triangle Triangle Mid Segment Theorem
Top
Triangle Mid Segment Theorem
Triangle is a most important two dimensional polygon in mathematics. In geometry, we deal with various geometric figures like rectangle, square, cube, parallelogram etc. Among all triangles are polygons with three vertices and three edges. Many practical problems are based on triangle theorems such as SideSplitter Theorem, Angle Bisector Theorem and Area Ratios of triangle.
Triangle mid segment is one of the basic theorems on triangles which explains a very important property of triangles. Before working on midsegment theorem first we should know about “midsegment”, A midsegment of a triangle connects the midpoints of two sides of a triangle. In this section will study about triangle mid segment theorem in detail.
Related Calculators  
Mid Point Calculator  Mid Quartile Calculator 
Bayes Theorem Calculator  Binomial Theorem Calculator 
Proof
Back to Top
Statement: A segment that joins the mid points of any two sides is measure half of the length of third side and also parallel to it.
That is, ST = $\frac12$ QR and ST ll QR.
Proof:
Since S and T are mid points of triangle.
SP = QS and TP = RT
Also PR = 2PT and PQ = 2PS
Consider two triangles, $\triangle$PST and $\triangle$PQR,
$\fracPSPQ$ = $\fracPTPR$ = $\frac12$ (angle P is common)
Which shows that both the triangles are similar.
or $\triangle$PST $\sim$ $\triangle PQR$
Since ST and QR are corresponding parts of the similar triangles, which implies
$ \fracPSPQ$ = $\fracPTPR$ = $\fracSTQR$ = $\frac12$
or $\fracSTQR$ = $\frac12$
or ST = $\frac12$ (QR)
Also $\angle$ PTS = $\angle$ PRQ (corresponding angles of similar triangles)
Transversal line PR cuts line segments ST and QR (By transversal property)
Using above results, we obtained
ST ll QR
Hence Proved.
Converse
Back to Top
Statement: The straight line passing through the midpoint of one leg of a triangle bisect another leg at its midpoint.
Say S is the midpoint of PQ
To prove: T is midpoint of PR
Extends line passing through S till V and join RV in such a way
QS  RV and SV  QR
Proof:
In parallelogram QSVR,
QS = RV
Also PS = QS (S is mid point of PQ)
=> RV = PS …(1)
In $\triangle$ PST and $\triangle$ RTV:
$\angle P$ = $\angleR$ (RV QS) …(2)
$\anglePTS$ = $\angleRTV$ (vertical opposite angles) …(3)
From (1), (2) and (3)
$\triangle$ PST $\cong$ $\triangle$ RTV
=> PT = TR
=> T is midpoint of PR.
Examples
Back to Top
Below are some examples on mid segment theorem:
Example 1: If the vertices of any triangle are (3, 5), (1, 3) and (5, 2). Show that mid segment is half the length to the side opposite to it.
Solution: Consider a triangle ABC with vertex coordinates A(3, 5), B(1, 3) and C (5, 2).
Step 1: Find the midpoints of two sides
Consider E and F are midpoints of AB and AC respectively
Midpoint of AB = $(\frac3+12, \frac5+32)$ = (2, 4) or E(2, 4)
Midpoint of AC = $(\frac3+52, \frac522)$ = (4, 1.5) or F(4, 1.5)
Step 2: Measure of EF and BC (using distance formula)
EF = $\sqrt(42)^2 + (1.5 – 4)^2$ = $\sqrt4 + 6.25$ = $\sqrt10.25$ = 3.202
BC = $\sqrt(51)^2 + (2 – 3)^2$ = $\sqrt16 + 25$ = $\sqrt41$ = 6.403
Step 3:
EF = $\frac12$ (BC) = $\frac12$ $\times$ 6.403 = 3.202 (True, from step 2)
Example 2: Find the value of m. ST is parallel to RQ.
Solution: Since mid segment of a triangle is half the length of its parallel side.
That is, ST = $\frac12$(RQ)
ST = m and RQ = 30 (given)
Substitute the number into above equation
=> m = $\frac12$ $\times$ 30
or m = 15
The value of m is 15 units.
Example 3: In $\triangle$ XYZ, if R is the midpoint of XY and S is the midpoint of XZ. Also RS  YZ. Find the
lengths of RS and YZ if RS = x – 3 and YZ = 4x – 16.
Solution: Since RS  YZ
RS = x – 3 and YZ = 4x – 16 (given)
Using mid segment theorem, we get
x – 3 = $\frac12$ (4x – 16)
x – 3 = 2x – 8
x = 5
Now RS = x – 3 = 5 – 3 = 2 and
YZ = 4x – 16 = 20 – 16 = 4
=> RS = 2 and YZ = 4
Therefore, length of RS is 2 units and length of YZ is 4 units.
Related Topics  
Math Help Online  Online Math Tutor 
 Triangle Mid Segment Theorem
 Proof
 Converse
 Examples
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Contact Us 
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Homework Help 
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Sales Jobs 
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Answers 
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