Math, asked by sumitt6315, 8 months ago

ABC is a triangle in which ∠B=2∠C. D is a point on BC such that AD bisects ∠BAC and AB = CD. Prove that ∠BAC = 72°.

Answers

Answered by nikitasingh79
10

Given: ABC is a triangle in which ∠B = 2∠C. D is a point on BC such that AD bisects ∠BAC and AB = CD.

To Prove : ∠BAC = 72°.

 

Proof :

Construct the angular bisector of ∠ABC, which meets AC in P and join PD.

Let, ∠ACB = y

∠B = ∠ABC = 2∠C = 2y

Let ∠BAD = ∠CAD = x

∠BAC = 2x  

[ AD is the bisector of ∠BAC]

In ∆BPC, we have,  

∠CBP = ∠BCP = y  

[ BP is the bisector of ∠ABC)

Therefore, PC = BP

In ∆ABP & ∆DCP ,  

∠ABP = ∠DCP = y

AB = DC (Given)

PC = BP (Proved above)

So, by SAS congruence criterion , we obtain   ∆ABP ≅   ∆DCP

Now,

∠BAP = ∠CDP and, AP = DP (By CPCT)

∠BAP = ∠CDP = 2x and  ∠ADP = ∠DAP = x  

In  ∆ABD,  

∠ADC = ∠ABD + ∠BAD

[By exterior angle theroem ]

(∠ADP + ∠CPD) = ∠ABD + ∠BAD  

x + 2x = 2y + x  

x - x + 2x = 2y

2x = 2y

x = y …………(1)

In ∆ABC, we have,  

We know that, the sum of all angles of triangle is 180°.

∠A + ∠B + ∠C = 180°  

2x + 2y + y = 180°

[∠A = 2x, ∠B = 2y, ∠C = y]

2x + 2x + x = 180°

[From eq1 , x = y]

5x = 180°

x = 180°/5

x = 36°

∠BAC = 2x = 2 × 36°

∠BAC = 72°

Hence, ∠BAC = 72°.

HOPE THIS ANSWER WILL HELP YOU…..

 

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Attachments:
Answered by juveriaridha653
7

Answer:

Step-by-step explanation: Given: ABC is a triangle in which ∠B = 2∠C. D is a point on BC such that AD bisects ∠BAC and AB = CD.

To Prove : ∠BAC = 72°.

 

Proof :

Construct the angular bisector of ∠ABC, which meets AC in P and join PD.

Let, ∠ACB = y

∠B = ∠ABC = 2∠C = 2y

Let ∠BAD = ∠CAD = x

∠BAC = 2x  

[ AD is the bisector of ∠BAC]

In ∆BPC, we have,  

∠CBP = ∠BCP = y  

[ BP is the bisector of ∠ABC)

Therefore, PC = BP

In ∆ABP & ∆DCP ,  

∠ABP = ∠DCP = y

AB = DC (Given)

PC = BP (Proved above)

So, by SAS congruence criterion , we obtain   ∆ABP ≅   ∆DCP

Now,

∠BAP = ∠CDP and, AP = DP (By CPCT)

∠BAP = ∠CDP = 2x and  ∠ADP = ∠DAP = x  

In  ∆ABD,  

∠ADC = ∠ABD + ∠BAD

[By exterior angle theroem ]

(∠ADP + ∠CPD) = ∠ABD + ∠BAD  

x + 2x = 2y + x  

x - x + 2x = 2y

2x = 2y

x = y …………(1)

In ∆ABC, we have,  

We know that, the sum of all angles of triangle is 180°.

∠A + ∠B + ∠C = 180°  

2x + 2y + y = 180°

[∠A = 2x, ∠B = 2y, ∠C = y]

2x + 2x + x = 180°

[From eq1 , x = y]

5x = 180°

x = 180°/5

x = 36°

∠BAC = 2x = 2 × 36°

∠BAC = 72°

Hence, ∠BAC = 72°.

HOPE THIS ANSWER WILL HELP YOU…..

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