Question

Abc is a triangle in which angle b=angle c d is a mid point on side bc such that ad bisect angle bac and ab=cd prove that angle bac=72

Answer 1

In ΔABC, we have
∠B = 2∠C or, ∠B = 2y, where ∠C = y
AD is the bisector of ∠BAC. So, let ∠BAD = ∠CAD = x
Let BP be the bisector of ∠ABC. Join PD.
In ΔBPC, we have
∠CBP = ∠BCP = y ⇒ BP = PC ... (1)

Now, in ΔABP and ΔDCP, we have
∠ABP = ∠DCP = y
AB = DC [Given]
and, BP = PC [Using (1)]
So, by SAS congruence criterion, we have

In ΔABD, we have
∠ADC = ∠ABD + BAD ⇒ x + 2x = 2y + x ⇒ x = y
In ΔABC, we have
∠A + ∠B + ∠C = 180°
⇒ 2x + 2y + y = 180°
⇒ 5x = 180°
⇒ x = 36°
Hence, ∠BAC = 2x = 72°