Question

In the given figure, \(\triangle \mathrm{ABC}\) is isosceles and \(A P \times B Q=A C^{2}\), prove that \(\triangle A C P \sim \triangle B C Q\).

Answer 1

Given,

⇒ AP × BQ = AC²

⇒ AP × BQ = AC × AC

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

⇒ `(AP)/(BC) = (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)]

⇒ `(AP)/(BC) = (AC)/(BQ)`​ [From equation (1)]

∴ Δ ACP ∼ Δ BCQ (By SAS postulates)