Question

15. In a AABC, AB = AC and D is a point on AB,
such that AD = DC = BC. Then, ZBAC is
(a) 40°
(b) 45° (C) 30° (d) 36​

Answer 1

Answer:

I think the answer is (d) 36

Step-by-step explanation:

CD=BC

angle DBC = angle BDC = x

AB = AC

So, angle ABC = angle ACB = x

angle ABC+ angle BCA+ angle BAC = 180

angle BAC+ x+ x= 180

angle BAC= 180 - 2x

AD=CD

So, angle ACD = angle DAC

sO, angle DAC = 180-2x

angle BCD = angle ACB - angle ACD = x-(180-2x) = 3x-180

angle BDC+ angle DBC+ angle BCD = 180

x+ x+ 3x= 180

5x = 180

x = 72

angle BAC = 180-2x = 180-2(72) = 36

Answer 2

Answer:

attached pic

Step-by-step explanation:

As, angles opposite to the equal sides are always equal.

So, \angle DBC = \angle BDC = x∠DBC=∠BDC=x (Let)

As, AB = AC

So, \angle ABC = \angle ACB = x∠ABC=∠ACB=x

In triangle ABC,

By angle sum property of the triangle ABC, we get

\angle ABC+ \angle BCA+ \angle BAC = 180^\circ∠ABC+∠BCA+∠BAC=180

∘

\angle BAC+ x+ x= 180^\circ∠BAC+x+x=180

∘

\angle BAC= 180^\circ - 2x∠BAC=180

∘

−2x

As, AD=CD

So, \angle ACD = \angle DAC∠ACD=∠DAC

sO, \angle DAC = 180^\circ-2x∠DAC=180

∘

−2x

Now, consider \angle BCD = \angle ACB - \angle ACD = x-(180-2x) = 3x-180^\circ∠BCD=∠ACB−∠ACD=x−(180−2x)=3x−180

∘

By angle sum property of the triangle BCD, we get

\angle BDC+ \angle DBC+ \angle BCD = 180^\circ∠BDC+∠DBC+∠BCD=180

∘

x+ x+ 3x= 180^\circx+x+3x=180

∘

5x = 180

x = 72^\circx=72

∘

Now, \angle BAC = 180-2x = 180-2(72) = 36^\circ∠BAC=180−2x=180−2(72)=36

∘

Hence, Proved.

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