Question

Prove that $\Delta ABC$ is isosceles if any one of the following holds true:

• A. Altitude AD bisects $\angle BAC$
• B. Median AD is perpendicular to BC
• C. Both A and B
• D. None of these

Answer 1

The correct option is C Both A and B

A) $\underset{–}{AltitudeADbisects\angle BAC}$

In $△BAD$ and $△CAD$

$AD=AD$ (Common Side)

$\angle ADB=\angle ADC={90}^o$ (AD is altitude)

$\angle BAD=\angle CAD$ (AD bisects $\angle BAC$)

$\therefore △BAD\cong △CAD(A.S.A)$

$\Rightarrow AB=AC(C.P.C.T.C.)$

Thus, $△BAC$ is an isosceles triangle.

B) $\underset{–}{MedianADisperpendiculartoBC}$

In $△BAD$ and $△CAD$

$AD=AD$ (Common Side)

$\angle ADB=\angle ADC={90}^o$ (AD is perpendicular to BC)

$BD=CD$ (AD is Median)

$\therefore △BAD\cong △CAD(S.A.S.)$

$\Rightarrow AB=AC(C.P.C.T.C.)$

Thus, $△BAC$ is an isosceles triangle.