Question

∆ ABC is isosceles with AB = AC. Line segment AD bisects ∠A and meets the base BC in D.
(i) Is ∆ ADB ≅ ∆ ADC?
(ii) State the three pairs of matching parts used to answer (i).
(iii) Is it true to say that BD = DC?

Answer 1

(i) We have AB = AC (given)
$\angle BAD=\angle CAD$ (AD bisects $\angle BAC$)
and AD = AD (common)
Therefore by SAS condition of congruence, $△ABD\cong △ACD$.

(ii) We have used AB, AC; $\angle BAD=\angle CAD$; AD, DA.

(iii) Now$△ABD\cong △ACD$ therefore by c.p.c.t BD = DC.