ABCD is a trapezium in which AB || DC. P and Q are points on sides

Solution:

In trapezium ABCD, AB || DC. P and Q are points on sides AD and BC such that PQ || AB.

Join AC. Suppose AC intersects PQ in O.

In $\Delta A C D, O P \| C D$

$\therefore \frac{A P}{P D}=\frac{A O}{O C} \quad \ldots \ldots(1) \quad(\mathrm{BPT})$

In $\Delta A B C, O Q \| A B$

$\therefore \frac{B Q}{Q C}=\frac{A O}{O C} \quad \ldots . .(2) \quad(\mathrm{BPT})$

From (1) and (2), we get

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

$\frac{A P}{18}=\frac{35}{15}$

$A P=\frac{35 \times 18}{15}$

$A P=\frac{7 \times 5 \times 3 \times 6}{5 \times 3}$

$A P=42$

$A D=A P+P D$

 

$A D=42+18$

$A D=60$

Hence, the value of $A D$ is 60 .