Question

If D is the mid-point of the hypotenuse AC of a right triangle ABC, prove that BD = $\frac{1}{2}$AC. [4 MARKS]

Answer 1

Concept : 1 Mark
Process : 2 Marks
Proof : 1 Mark

To Prove: BD = $\frac{1}{2}$AC

Construction: Produce BD to E such that BD = DE and join EC

Proof: In $\Delta$ADB and $\Delta$CDE, we have

AD = DC [Given]

BD = DE [By Construction]

And, $\angle$ADB = $\angle$CDE [Vertically opposite angles]

So,

$△$ADB $\cong$ $△$CDE [by SAS criterion of congruence]

$\Rightarrow$ EC = AB and $\angle$CED = $\angle$ABD [C.P.C.T.C] ............(1)

But $\angle$CED and $\angle$ABD are alternate angles and are equal.

So, CE || AB

$\angle$ABC + $\angle$ECB =${180}^{\circ }$ [Co - interior angles angles]

90^o + $\angle$ECB = ${180}^{\circ }$

$\angle$ECB = ${90}^{\circ }$

In $\Delta$s ABC and ECB,

AB = EC [From 1]

BC = CB [Common]

And, $\angle$ABC = $\angle$ECB =${90}^{\circ }$

So

$△$ABC $\cong$ $△$ECB [by SAS criterion of congruence]

$\Rightarrow$ AC = BE [C.P.C.T.C]

$\frac{1}{2}$AC = $\frac{1}{2}$BE

$\therefore$ BD = $\frac{1}{2}$AC