Question

ABC is a triangle in which angle B= 2 angle C D is a point on side BC such that AD bisect angle BAC and AB==D. prove that angle BAC=72degree

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°

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Answer 2

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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