In an isosceles Δ ABC, the base AB is produced both the ways to P and Q such that AP × BQ = AC². Prove that Δ APC∼Δ BCQ.

Given: CA – CB

AP x BQ = AC²

To prove: ΔAPC ~ BCQ

Theorem Used:

If two corresponding sides and one angle of two triangles are equal, the triangles are said to be similar.

Proof:

AP X BQ = AC² (Given)

⇒ AP x BC = AC x AC

As AC = BC

AP x BC = AC x BC

…… (i)

Since, CA = CB (Given)

Then, ∠CAB = ∠CBA (Opposite angle to equal sides) …. (ii)

Now, ∠CAB +∠CAP = 180° (Linear pair of angles) …… (iii)

And ∠CBA + ∠CBQ = 180° (Linear pair of angles) ……. (iv)

Compare equation (ii) (iii) & (iv)

∠CAP = ∠CBQ …… (v)

In ΔAPC and ΔBCQ

∠CAP = ∠CBQ (From equation v)

(From equation i)

Then, ΔAPC ~ ΔBCQ (By SAS similarity)