In a triangle ABC, AB = AC and D is a point on side AC such that BC² = AC × CD.

Prove that BD = BC.

Draw BE ⊥ AC.

In right triangle BEC,

By pythagoras theorem,

⇒ BC² = BE² + EC²

⇒ BC² = BE² + (AC - AE)²

⇒ BC² = BE² + AC² + AE² - 2AC.AE ………(i)

In right triangle ABE,

By pythagoras theorem,

⇒ AB² = BE² + AE²………(ii)

Substituting value of BE² + AE² from (ii) in (i) we get,

⇒ BC² = AB² + AC² - 2AC.CE

⇒ BC² = AC² + AC² - 2AC.AE [∵ AB = AC]

⇒ BC² = 2AC² - 2AC.AE

⇒ BC² = 2AC(AC - AE)

⇒ BC² = 2AC × EC ………(iii)

Given, BC² = AC × CD ……….(iv)

Comparing (iii) and (iv) we get,

⇒ 2AC × EC = AC × CD

⇒ 2EC = CD or EC = $\dfrac{\text{CD}}{2}$

So, we can say that ED = EC as E is mid-point of CD.

In △BDE and △BCE,

BE = BE (Common)

ED = EC

∠BED = ∠BEC = 90°

△BDE ≅ △BCE by SAS axiom of congruency.

We know that corresponding parts of congruent triangle are equal.

∴ BD = BC.

Hence, proved that BD = BC.