Question

Prove the theorem : The segment joining midpoints of any two sides of a triangle is parallel to the third side and half of it.​

Answer 1

Answer:

The line segment joining the mid-points of two sides of a triangle is parallel to the third side. You can prove this theorem using the following clue: Observe the figure in which E and F are mid-points of AB and AC respectively and CD || BA. So, EF = DF and BE = AE = DC.

Step-by-step explanation:

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Answer 2

Statement

The segment joining midpoints of any two sides of a triangle is parallel to the third side and half of it.

Given

In triangle ABC, point P is the midpoint of seg AB and point A is the midpoint of seg AC

To prove

Seg PQ || seg BC

$\sf \: and \: PQ = \dfrac{1}{2} \: BC$

Construction

Produce seg PQ upto R such that PQ = QR

Draw seg RC.

Proof

In triangle AQP and triangle CQR

seg PQ ≅ seg QR. ..... (Construction )

seg AQ ≅ seg QC ....... ( given )

∠ AQP ≅ ∠ CQR............( vertically opposite angle )

Triangle AQP ≅ Traingle CQR....( SAS test )

∠PAQ ≅ ∠RCQ.... (1) c.a.c.t

therefore,

seg AP ≅ seg CR..... (2) c.s.c.t

From (1) line AB || line CR... Alternate angle test

From (2) seg AP ≅ seg CR

Now, seg AP ≅ seg PB ≅ CR and seg PB || seg CR

□ PBCR is a parallelogram.

seg PQ || seg BC and PR = BC... (opposite sides are congruent )

$\sf \: PQ = \dfrac{1}{2} \: PR.... ( Construction )$

$\sf \: PQ = \dfrac {1}{2} \: BC$

PR = BC