Question

AD is an altitude of an isosceles triangle ABC in which AB = AC.
Show that, (i) AD bisects BC (ii) AD bisects ∠ A.

Answer 1

Answer:

Step-by-step explanation:

(i) In ΔBAD and ΔCAD ,

 AB = AC ( GIVEN )

AD = AD ( COMMON )

∠ ADB = ∠ ADC ( AD ⊥ BC )

∴  Δ BAD ≅ Δ CAD ( RHS CONGRUENCE RULE )

⇒ BD = CD ( CPCT )

HENCE AD BISECTS BC

(ii)  ALSO BY CPCT ,

∠ BAD = ∠CAD

HENCE AD BISECTS ∠ A

please refer to the below image