Question

In the given figure, AB = AC and D is the midpoint of BC
(i) State the three pairs of equal parts in ΔADB and ΔADC.
(ii) Is ΔADB $\cong$ ΔADC? Give reasons.
(iii) Is ∠B = ∠C? Why?

Answer 1

In $△ABC$,
AB = AC
D is the midpoint of BC
BD = DC
Now in △ADB and △ADC
AB = AC (Given)
AD = AD (Common)
BD = DC (D is the midpoint of BC)
(ii) △ADB $\cong$△ADC (SSS axiom)
(iii) ∠B = ∠C( by c.p.c.t).