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Question
In the figure, given below, straight lines AB and CD intersect at P; and AC || BD. Prove that: If BD = 2.4 cm, AC = 3.6 cm, PD = 4.0 cm and PB = 3.2 cm; find the lengths of PA and PC.
Solution
In ΔAPC and ΔBPD,
∠APC = ∠BPD ...(Vertically opposite angles)
∠ACP = ∠BDP ...(Alternate angles since AC || BD)
∴ ΔAPC ∼ ΔBPD ...(AA criterion for similarity)
So, `(PA)/(PB) = (PC)/(PD) = (AC)/(BD)`
`=> (PA)/(3.2) = (PC)/4 = 3.6/2.4`
So, `(PA)/(3.2) = (3.6)/(2.4)` and `(PC)/4 = (3.6)/(2.4)`
`=> PA = (3.6 xx 3.2)/2.4 = 4.8 cm`
And PC = `(3.6 xx 4)/(2.4)` = 6 cm
Hence, PA = 4.8 cm and PC = 6 cm
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