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Question
□ABCD is a parallelogram. Point P is the midpoint of side CD. seg BP intersects diagonal AC at point X, then prove that: 3AX = 2AC
Solution
Proof: From the figure, in ΔABX and ΔCPX
As, AB || CD
∠BAX = ∠PCX ......[Alternate angle]
∠BXA = ∠PXC ........[Vertically opposite angles]
∴ ΔABX ∼ ΔCPX .......[By AA similarity theorem]
We know that,
Similar triangles have comparable side ratios that are similar to or equal.
∴ `("AX")/("CX") = ("AB")/("CP")`
But CD = AB and 'P' is mid-point of CD.
∴ AB = 2CP
∴ `("AX")/(("AC" - "AX")) = (2"CP")/("CP")`
∴ `("AX")/(("AC" - "AX"))` = 2
∴ AX = 2(AC – AX)
∴ AX = 2AC – 2AX
∴ AX + 2AX = 2AC
∴ 3AX = 2AC
Hence proved.
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