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

Answer 1

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.