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Standard IX

Mathematics

Intercept Theorem

In a parallel...

Question

In a parallelogram ABCD, the line drawn through a point P on AB; parallel to BC, meets AC at Q. The line through Q, parallel to AB meets AD at R. Then, prove that $\frac{A P}{P B} =$ $\frac{A Q}{Q C} =$ $\frac{A R}{R D}$ .

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Solution

In $Δ A C D,$

QR || CD

$\frac{A Q}{Q C} = \frac{A R}{R D}$ ..............(i)

In $Δ A C B,$

PQ || CB

$\frac{A Q}{Q C} = \frac{A P}{P B}$ ..............(i)

On comparing (i) & (ii), we get

$\frac{A P}{P B} = \frac{A R}{R D}$ ..............(iii)

Using result (i) & (ii) & (iii)

$\frac{A P}{P B} = \frac{A Q}{Q C} = \frac{A R}{R D}$

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In a parallelogram ABCD, the line drawn through a point P on AB; parallel to BC, meets AC at Q. The line through Q, parallel to AB meets AD at R. Then, $\frac{A P}{P B} =$

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