Question

In the figure CD is the angle bisector of angle C, angle B = angle ACE. Prove that angle ADC = angle ACD

Answer 1

Ad is equal to ac (isoceles triangle's equal sides)
So if sides of isosceles triangle are equal so angles are also equal so angle ADC is equal to angle ACD
Hence proved

Answer 2

Given:

CD is the bisector of angle C

Angle B = Angle ACE

To find:

Prove that angle ADC = angle ACD

Solution:

In ΔBCD, we have

∠BCD + ∠BDC + ∠DBC = 180° ....... [Angle sum property] .... (i)

Also, ∠ADC + ∠BDC = 180° ..... [Linear Pair] ...... (ii)

From equation (i) & (ii), we get

∠BCD + ∠BDC + ∠DBC = ∠ADC + ∠BDC

⇒ ∠BCD + ∠BDC + ∠DBC - ∠BDC = ∠ADC

⇒ ∠BCD + ∠DBC = ∠ADC ............... (iii)

Now, according to the question we have,

∠ABC = ∠ACE ...... (iv)

and

∠BCD = ∠ECD  .... [∵ CD is the bisector of angle C as given in the figure]...(v)

On adding eq. (iv) & (v), we get

∠ABC + ∠BCD = ∠ACE + ∠ECD

⇒ ∠ABC + ∠BCD = ∠ACD ...... (vi)

On comparing eq. (iii) & (vi), we get

$\boxed{\boxed{\bold{\angle ADC = \angle ACD}}}$

Hence proved

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