Question

In the adjoining figure, AB = BC and AC = CD. Prove that ∠BAD : ∠ADB = 3 : 1.

Answer 1

Given:

• B, C, D are collinear with C between B and D.
• AB = BC and AC = CD.

To Prove:

• ∠BAD : ∠ADB = 3 : 1.

Proof [Step-wise]:

1. Let x = ∠ADC.

Since AC = CD, triangle ACD is isosceles.

So, base angles at A and D are equal: ∠CAD = ∠ADC = x. 

Isosceles triangle → Equal base angles.

2. Since AB = BC, triangle ABC is isosceles.

So, its base angles are equal: set y = ∠BAC = ∠ACB.

Again, using the isosceles-triangle property.

3. Consider point C on the straight line BD.

The two angles formed at C by ray CA with the two directions CB and CD are supplementary.

So, ∠ACB + ∠ACD = 180°.

4. But in triangle ACD, 

∠ACD = 180° – (∠CAD + ∠ADC)

= 180° – (x + x)

= 180° – 2x

Substitute into the relation from step 3:

y + (180° – 2x) = 180°

Hence, y = 2x.

5. Now ∠BAD = ∠BAC + ∠CAD 

= y + x

= 2x + x

= 3x

While ∠ADB = ∠ADC = x.

6. Therefore, ∠BAD : ∠ADB

= 3x : x

= 3 : 1