Question

In the given figure; $\angle 1=\angle 2$ and AB = AC. Prove that: [4 MARKS]



(i) $\angle B=\angle C$

(ii) BD = DC

(iii) AD is perpendicular to BC.

Answer 1

Steps: 1 Mark
Each proof: 1 Mark

In $\Delta ADB$ and $\Delta ADC$

$\angle 1=\angle 2$ [Given]

$\Rightarrow \angle BAD=\angle CAD$

$AD=AD$ [Common side]

$AB=AC$ [Given]

$\Rightarrow \Delta ADB\cong \Delta ADC$ [SAS congruency criteria]

(i) $\therefore \angle B=\angle C$ [Corresponding parts of congruent triangles]

(ii) $BD=DC$ [Corresponding parts of congruent triangles]

(iii) $\Delta ADB\cong \Delta ADC$ [proved above]

$\angle ADB+\angle ADC=180°$ [Linear pair]

$\Rightarrow \angle ADB=\angle ADC$ [c.p.c.t]

$\therefore \angle ADB+\angle ADB=180°$

$\Rightarrow 2\angle ADB=180°$

$\Rightarrow \angle ADB=\frac{180°}{2}$

$=90°$

$\Rightarrow AD\perp BC$