Question

In ABC, AD is angle bisector,
angle BAC = 111 and AB+BD=AC find the value of angle ACB=?​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Given :-

• AD is angle bisector .
• ∠BAC = 111°
• AB + BD = AC .

To Find :-

• ∠ACB = ?

Construction :- (Refer to image .)

• Extend B to Point E such that, BE = BD .
• Also, ED = DC .

Solution :-

in ∆BED, we have :-

→ BE = BD ( By construction.)

Than,

→ ∠BED = ∠BDE = θ . { Angles opposite to equal sides of a triangle are equal.}

Now,

→ ∠ABD = ∠BED + ∠BDE . { Exterior angle is equal to sum of two opposite interior angles.}

→ ∠ABD = θ + θ

→ ∠ABD = 2θ .

Now, From Second Diagram of image , we get :-

→ AB + BD = AC (given)

→ BD = BE (By consturction .)

→ AE = AB + BE

So,

→ AE = AB + BD

∴

→ AE = AC .

Now, in ∆AEC, we have :-

in ∆AED and ∆ACD ,

→ AD = AD .(common.)

→ ∠EAD = ∠CAD . {AD is angle bisector.}

→ AE = AC. (Proved above. )

So,

→ ∆AED ~ ∆ACD ( By SAS. )

Than,

→ ∠AED = ∠ACD ( By CPCT.)

Therefore,

→ ∠ACD = θ.

Now, in ∆ABC we have :-

→ ∠ACB = θ

→ ∠ABC = 2θ.

→ ∠BAC = 111° .(given)

So,

→ ∠ACB + ∠ABC + ∠BAC = 180° . (Angle sum Property.)

→ θ + 2θ + 111° = 180°

→ 3θ = 180° - 111°

→ 3θ = 69°

→ θ = 23°.

Hence,

→ ∠ACB = 23° (Ans.)

∴ ∠ACB will be 23° .