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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.
Solution
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.
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