In a triangle ABC, if AB = AC and AB is produced to D such that BD = BC, find ∠ACD: ∠ADC.

#### Solution

In the given ΔABC,AB = ACand *AB *is produced to *D* such that BD = BC

We need to find ∠ACD : ∠ADC

Now, using the property, “angles opposite to equal sides are equal”

As AB = AC

∠6 = ∠4 ........(1)

Similarly,

As BD = BC

∠1 = ∠2 ........(2)

Also, using the property, “an exterior angle of the triangle is equal to the sum of the two opposite interior angle”

In Δ*BDC*

ext. ∠6 = ∠1 + ∠2

ext. ∠6 = ∠1 + ∠1 (Using 2)

ext. ∠6 = 2∠1

From (1), we get

ext. ∠4 = ∠2 .......(3)

Now, we need to find ∠ACD : ∠ADC

That is,

(∠4 + ∠2): ∠1

(2∠1 + ∠2) : ∠1 (Using 3)

(2∠1 + ∠1) : ∠1(Using 2)

3∠1 :∠1

Eliminating ∠1from both the sides, we get 3:1

Thus, the ratio of ∠ACD :∠ADC is 3 :1