In the given figure, Δ ABC is isosceles and AP x BQ = AC², prove that Δ ACP ∼ Δ BCQ.

Given,

⇒ AP x BQ = AC²

⇒ AP x BQ = AC x AC

⇒ AP x BQ = AC x BC (From figure, AC = BC)

⇒ $\dfrac{AP}{BC} = \dfrac{AC}{BQ}$ ……………………(1)

Since, AC = BC

⇒ ∠CAB = ∠CBA ……………(2) [Angles opposite to equal sides are equal]

⇒ 180° - ∠CAB = 180° - ∠CBA

⇒ ∠CAP = ∠CBQ ……………….(3)

In Δ ACP and Δ BCQ,

⇒ ∠CAP = ∠CBQ [From equation (3)]

⇒ $\dfrac{AP}{BC} = \dfrac{AC}{BQ}$ [From equation (1)]

∴ Δ ACP ∼ Δ BCQ (By SAS postulates)

Hence, proved that Δ ACP ∼ Δ BCQ.