The side BC of a triangle ABC is produced to D so that CD = AC. If the angle BAD = 109° and the angle ACB = 72°, prove that BD is greater than AD.

Given: ∠BAD = 109°, ∠ACB = 72° and CD = AC (i.e., Δ ACD is isosceles).

∠BCA + ∠ACD = 180° (∴ Linear pair)

⇒ 72° + ∠ACD = 180°

⇒ ∠ACD = 180° - 72°

⇒ ∠ACD = 108°

Since AC = CD, Δ ACD is isosceles. So:

⇒ ∠ADC = ∠CAD

Using the angle sum property in Δ ACD;

∠ACD + ∠ADC + ∠CAD = 180°

⇒ 108° + ∠ADC + ∠CAD = 180°

⇒ 2∠ADC = 180° - 108°

⇒ 2∠ADC = 72°

⇒ ∠ADC = $\dfrac{72°}{2}$ = 36° and ∠CAD = 36°

From the figure, ∠BAD = ∠BAC + ∠CAD

⇒ 109° = ∠BAC + 36°

⇒ ∠BAC = 109° - 36° = 73°

In Δ ABC,

∠ABC + ∠ACB + ∠CAB = 180°

⇒ ∠ABC + 72° + 73° = 180°

⇒ ∠ABC + 145° = 180°

⇒ ∠ABC = 180° - 145°

⇒ ∠ABC = 35°

In Δ ABD, the angles are ∠A = 109°, ∠B = 35° and ∠D = 36°.

Since larger angles correspond to larger sides, we observe:

∠A > ∠D > ∠B

So, the corresponding sides follow:

⇒ BD > AB > AD

Hence, BD is greater than AD.