Question

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

Answer 1

In the given triangle ABC, AB=AC and BD=BC.

• Let ∠ABC=x° and let ∠ADC=y°
• Since triangle ABC is isosceles with AB=AC, therefore ∠ABC=∠ACB=x°.
• Also, in triangle BDC, as BD=BC, ∠BDC=∠BCD=y°.
• In triangle BDC, ∠CBD=180°-∠BCD-∠BDC. Hence, ∠CBD=180°-2x.
• Now, since ABD is a straight line, ∠ABC + ∠CBD = 180°, or                        x + 180-2y =180. On solving, x=2y.
• Now, ∠ACD:∠ADC = (y+x) : y  =  3y : y  =  3

Hence we calculated the ratio of two angles as 3.

Answer 2

∠ACD : ∠ADC is 3 : 1

Step-by-step explanation:

The image given in the question is attached below.

From diagram, AB = AC and BD = BC

Now, on using the property, angles opposite to equal sides are equal.

∠AB = ∠AC

⇒ ∠6 = ∠4 → (1)

∠BD = ∠BC

⇒ ∠1 = ∠2 → (2)

On using the property 'an exterior angle of the triangle is equal to the sum of the two opposite interior angle', we get,

In ΔBDC

ext ∠6 = ∠1 + ∠2

ext ∠6 = ∠1 + ∠1 (from 2)

ext ∠6 = 2 ∠1

ext ∠4 = ∠2 (from 1) → (3)

Now, ∠ACD : ∠ADC

⇒ (∠4 + ∠2) : ∠1

(2 ∠1 + ∠2) : ∠1 (from 3)

(2 ∠1 + ∠1) : ∠1 (from 2)

3 ∠1 : ∠1

∴ 3 : 1 = ∠ACD : ∠ADC