Question

ABCD is a parallelogram whose sides AB and BC are 18cm and 12cm respectively. G is a point on AC such that CG : GA = 3 : 5 BG is produced to meet CD at Q and AD produced at P. Prove that ΔCGB ∼ ΔAGP. Hence, fi AP.

Answer 1

In ΔCGB and ΔAGP
∠CGB = ∠AGP   ...(vertically opposite angles)
∠GAP = ∠GCB  ...(AD || BC, therefore alternate angles)
Therefore, ΔCGB ∼ ΔAGP  ...(AA axiom)

∴ `"CG"/"GA" = "BC"/"AP"`

⇒ `(3)/(5) = (12)/"AP"`

⇒ AP = `(5 xx 12)/(3)`
⇒ AP = 20cm.