Question

Date Page Q3.- In A ABC, LB = 24C, D is a point on BC such that AD bisects LBAC and AB= CD. PT LBAC=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

CORRECT QUESTION

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

ANSWER

∠BAC = 72°

GIVEN

ABC is a triangle.

∠B = 2∠C

AD is the angle bisector of ∠BAC

AB = CD

TO FIND

To prove that ∠BAC = 72°

SOLUTION

We can simply solve the above problem as follows,

In ∆ABC,

∠B = 2∠C

Let,

∠B = 2y

∠C = y

AD is the angle bisector of ∠BAC, Ie., it divides the angle in two equal halves.

Let, ∠BAD = ∠CAD = x

and, ∠A= 2x

Let, BE be the angle bisector of ∠ABC, join DE.

Now,

In ∆BEC

∠CBE = ∠BCE = y

We know that if two angles in a triangle are equal, then the sides opposite to that side are also equal.

So,

BE = EC

In ∆'s ABE and DCE,

∠ABE = ∠DCE = y

It is given,

AB = DC

and, BE = EC

So, By SAS Congruence rule,

∆ABE ≅ ∆ DCE

∠BAE = ∠CDE

And, AE= DE

(∠A = 2x)

∠CDE = 2x and ∠ADE = ∠DAP = x

In ∆ABD

∠ADC = ∠ABD + ∠BAD

= x + 2x = 2y + x

So,

x= y

in ∆ABC

∠A + ∠B + ∠C = 180°

Putting the values, we have -

2x + 2y + y = 180°

Since, x = y

5x = 180°

x = 36.

So, ∠BAC = 2x = 2× 36= 72

Hence, ∠BAC = 72°. proved.

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