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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}.

In ΔABC, we have

∠B = 2∠C or, ∠B = 2*y*, 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* + 2*x* = 2*y* + *x* ⇒ *x* = *y*

In ΔABC, we have

∠A + ∠B + ∠C = 180°

⇒ 2*x* + 2*y* + *y* = 180°

⇒ 5*x* = 180°

⇒ *x* = 36°

Hence, ∠BAC = 2*x* = 72°

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