Two straight lines parallel to the sides BC and BA respectively through two vertices A and C of ΔABC meet at D. Let’s prove that ∠ABC = ∠ADC.

Given: ΔABC, line l||BC, line m||BA, line l and line m are meeting at point D

To prove: ∠ABC = ∠ADC

The figure for the given question is as shown below,

Now in the given figure BA is parallel to line m with AC as tranversal line, so

∠BAC = ∠ACD……….(i) (as they form is alternate interior angles)

Similarly, BC is parallel to line l with AC as tranversal line, so

∠ACB = ∠CAD……….(ii) (as they form is alternate interior angles)

Now consider ΔABC,

We know in a triangle the sum of all three interior angles is equal to 180°.

So in this case,

∠ABC + ∠BAC + ∠ACB = 180°

Substituting the values from equation(i) and (ii), we get

∠ABC + ∠ACD + ∠CAD = 180°

⇒ ∠ABC = 180° - ∠ACD - ∠CAD…………(iii)

Now consider ΔADC,

We know in a triangle the sum of all three interior angles is equal to 180°.

So in this case,

∠ADC + ∠CAD + ∠ACD = 180°

⇒ ∠ADC = 180° - ∠CAD - ∠ACD…..(iv)

Equating equation (iii) and equation (iv), we get

∠ABC = ∠ADC

Hence proved