abc is a triangle with angle b = 2(angle c). d is a point on bc such that ad bisects angle bac and ab = cd.Prove angle bac = 72 degree

Dear student,
The solution to question 4 is given below:

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°

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