Q.9. Plz:

9. ABC is a triangle in which $\angle$B= 2 $\angle$C. D is a point on BC such that AD bisects $\angle$BAC and AB=CD. Prove that $\angle$BAC=72^o.
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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°