Question

In the given figure, $AD$ bisects $\angle BAC,AB=AC$. Show that

(i) $∆ABD\cong ∆ACD$

(ii) $AD$ is perpendicular to $BC$

(iii) $AD$ bisects $BC$.

Answer 1

STEP 1 : Proving that $∆ABD\cong ∆ACD$

(i)

In $∆ABD$ and $∆ACD$,

$AB=AC$ [ Given ]

$\angle BAD=\angle CAD$ [ $AD$ bisects $\angle BAC$]

$AD=AD$ [ Common side ]

Therefore, by SAS congruence condition $∆ABD\cong ∆ACD$

Hence proved.

STEP 2 : Proving that $AD$ is perpendicular to $BC$

(ii)

Since, $∆ABD\cong ∆ACD$

As corresponding parts of congruent triangles are equal.

$\therefore \angle ADB=\angle ADC...\left(i\right)$ [ By C.P.C.T.]

Also, $\angle ADB+\angle ADC=180°...\left(ii\right)$ [ Linear Pair as $BDC$ is a straight line ]

$\Rightarrow \angle ADC+\angle ADC=180°$ [ Using equation (i) ]

$\Rightarrow 2\angle ADC=180°$

$\Rightarrow \angle ADC=\frac{180°}{2}=90°$

Since, $\angle ADB=\angle ADC=90°$

Hence, $AD$ is perpendicular to $BC$.

STEP 3 : Proving that $AD$ is bisects $BC$

(iii)

Since, $∆ABD\cong ∆ACD$

As corresponding parts of congruent triangles are equal.

$\therefore BD=CD$ [ By C.P.C.T.]

This implies that, $D$ is the mid point of $BC$

Hence, $AD$ bisects $BC$.