Question

ABC is a triangle where angle B= 2 x angle C and AD is a line on BC and bisects angle A. Prove that angle BAC = 72°

Answer 1

Step-by-step explanation:

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

In Δ′s ABP and DCP, we have

∠ABP=∠DCP, we have

∠ABP=∠DCP=y

AB=DC     [Given]

and, BP=PC     [As proved above]

So, by SAS congruence criterion, we obtain

ΔABP≅ΔDCP

⇒∠BAP=∠CDP and AP=DP

⇒∠CDP=2x and ∠ADP=DAP=x     [∴∠A=2x]

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=y]

⇒x=36∘

Hence, ∠BAC=2x=72∘

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

⇒BP=PC

In Δs, ABP and DCP, we have

∠ABP=∠DCP, we have

∠ABP=∠DCP=y

AB=DC [Given]

and, BP=PC [As proved above]

So, by SAS congruence criterion, we obtain

ΔABP≅ΔDCP

⇒∠BAP=∠CDP and AP=DP

⇒∠CDP=2x and ∠ADP=DAP=x [∴∠A=2x]

In ΔABD, we have

∠ADC=∠ABD+∠BAD

⇒x+2x=2y+x

⇒x=y

In ΔABC, we have

∠A+∠B+∠C=180⁰

⇒2x+2y+y=180⁰

⇒5x=180 ⁰ [∵x=y]

⇒x=36 ⁰

Hence, ∠BAC=2x =72⁰