Question

ΔABC is an isosceles triangle with AB = AC. Another triangle BDC is drawn with base BC = BD in such a way that BC bisects ∠B. If the measure of ∠BDC is 70°, find the measures of ∠DBC and ∠BAC.

Answer 1

In ΔBDC,
∠BDC = 70°
BD = BC
Therefore, ∠BDC = ∠BCD
⇒ ∠BCD = 70°
Now ∠BCD + ∠BDC + ∠DBC = 180°
70° + 70° + ∠DBC = 180°
∠DBC = 40°
∠DBC =∠ABC  ...(BC is the angle bisector)
⇒ ∠ABC = 40°
In ΔABC,
Since AB = AC, ∠ABC = ∠ACB ⇒ ∠ACB = 40°
∠ACB + ∠ABC + ∠BAC = 180°
40° + 40° + ∠BAC = 180°
∠BAC = 100°
Hence, ∠BAC = 100° and ∠DBC = 40°.