Question

State and prove midpoint theorem.

Answer 1

Statement of midpoint theorem

The midpoint theorem states that the line segment in a triangle joining the midpoint of two sides of the triangle is parallel to its third side and is also half of the length of the third side.

Construction: Draw CF parallel to AB such that DE extended to join at F.

Proof:

Step 1: Verify that $△$ADE$\cong$$△$CEF

Now, consider $△$ADE and $△$CEF,

$\angle$DAE$=$$\angle$FCE ( alternate angles as CF$\parallel$AB)

AE$=$CE ( E is the mid point of AC)

$\angle$AED$=$$\angle$CEF ( vertically opposite angles)

By (ASA), $△$ADE$\cong$$△$CEF

Step 2: Verify that BDFC is a parallelogram

DE$=$FE ( By C.P.C.T) $\dots \dots .\left(1\right)$

AD$=$CF ( By C.P.C.T)

But AD$=$BD, so, BD$=$CF

We have now, BD$=$CF, BD$\parallel$CF. One pair of the opposite sides is equal and parallel, therefore, BDFC is a parallelogram.

Step3: Use property of parallelogram to verify ,DE$\parallel$BC, DE$=$$\frac{1}{2}$BC

$\because$ BDFC is a parallelogram.

$\therefore$DE$\parallel$BC

$\because$ In a parallelogram opposite sides are equal.

$\therefore$DF$=$BC

$\Rightarrow$$2$DE$=$BC [Using (1) ]

$\Rightarrow$DE$=$$\frac{1}{2}$BC.

Hence, DE$\parallel$BC, and DE$=$$\frac{1}{2}$BC.

Hence, proved.