The given figure shows a trapezium in which AB is parallel to DC and diagonals AC and BD intersect at point P. If AP : CP = 3 : 5.

Find :

(i) △APB : △CPB

(ii) △DPC : △APB

(iii) △ADP : △APB

(iv) △APB : △ADB

(i) Since, △APB and △CPB have common vertex at B and their bases AP and PC are along the same straight line AC.

∴ △APB : △CPB = AP : PC = 3 : 5.

Hence, △APB : △CPB = 3 : 5.

(ii) In △DPC and △APB,

∠DPC = ∠APB [Vertically opposite angles are equal]

∠PAB = ∠PCD [Alternate angles are equal]

∴ △DPC ~ △APB.

We know that,

The ratio of the areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

$\therefore \dfrac{\text{Area of △DPC}}{\text{Area of △APB}} = \dfrac{PC^2}{AP^2}\\[1em] = \dfrac{5^2}{3^2} = \dfrac{25}{9}.$

Hence, △DPC : △APB = 25 : 9.

(iii) Since, △DPC ~ △APB,

$\therefore \dfrac{PD}{PB} = \dfrac{PC}{AP} = \dfrac{5}{3}$ ……….(1)

Since, △ADP and △APB have common vertex at A and their bases DP and PB are along the same straight line.

∴ △ADP : △APB = DP : PB = 5 : 3 ………[From 1]

Hence, △ADP : △APB = 5 : 3.

(iv) From part (iii)

$\dfrac{PD}{PB} = \dfrac{5}{3}$

Let PD = 5x and PB = 3x.

From figure,

BD = PD + PB = 5x + 3x = 8x.

Since, △APB and △ADB have common vertex at A and their bases PB and DB are along the same straight line.

∴ △APB : △ADB = PB : BD = 3x : 8x = 3 : 8

Hence, △APB : △ADB = 3 : 8.