Question

In the given figure, AB = AC, BC = CD and ∠A=40°, find ∠ACD

Answer 1

AB = AC
∴ ∠ABC = ∠ACB (Angles opposite to equal sides)
But ∠ABC + ∠ACB + ∠BAC = 180° (Angles of a triangle)
⇒ ∠ABC + ∠ABC + 40° = 180°
⇒ 2∠ABC = 180° − 40° = 140°
∴ ∠ABC = $\frac{{140}^{\circ }}{2}=70°$
In ΔBDC, BC = CD
∴ ∠ABC = ∠BDC (Each 70°)
In ΔADC, Ext. ∠BDC = ∠BAC + ∠DCA
40° + x = 70°
x = 30°