Question

ABC is an isosceles triangle with $AB=AC$. Prove: [4 MARKS]

(i) $\Delta ADB\cong \Delta ADC$

(ii) $\angle BAD=\angle CAD$

(iii) $BD=CD$

Answer 1

Properties: 1 Mark
Each proof: 1 Mark

In $\Delta ADBand\Delta ADC$

$AB=AC$ [Given]

$\angle ADB=\angle ADC={90}^{\circ }$ [Given]

$AD=AD$ [common]

Hence, $\Delta ADB\cong \Delta ADC$ [By RHS congruence rule…….(1)]

From (1), $\angle BAD=\angle CAD$ [Corresponding parts of congruent triangles]

From (1), $BD=DC$ [Corresponding parts of congruent triangles]