ABCD is a parallelogram with diagonal AC. If a line XZ is drawn such that XZ∥AB and cuts AC at Y, prove that  $\frac{B X}{X C}$ =  $\frac{A Z}{Z D}$ . [3 Marks]

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Solution

Given: AB || XZ

In ΔABC, AB || XY

Therefore $\frac{B X}{X C}$ = $\frac{A Y}{Y C}$ ….. (1) [1 Mark]

In parallelogram ABCD, AB || CD

AB || CD || XZ

In ΔACD, CD || YZ.

Therefore $\frac{A Y}{Y C}$ = $\frac{A Z}{Z D}$ ….(2) [1 Mark]

From (1) and (2)

$\frac{B X}{X C}$ = $\frac{A Y}{Y C}$ = $\frac{A Z}{Z D}$

So, $\frac{B X}{X C}$ = $\frac{A Z}{Z D}$ [1 Mark]

Hence proved.

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