ABCD is trapezium with AB || DC. The diagonal AC and BD intersect at E . If `Delta AED ~ Delta BEC` . Prove that AD = BC .

Options

A. `((p^(2) + q^(2) sin theta)/(p cos theta + q sin theta)`
B. `((p^(2) + q^(2)) cos theta)/(p cos theta + q sin thet)`
C. `(p^(2) + q^(2))/(p^(2) cos theta + q^(2) sin theta)`
D. `((p^(2) + q^(2)) sin theta)/((p cos theta + q sin theta))`

Answer

A

Explanation

`BD = sqrt(p^(2) + q^(2))`
`angleABD = angleBDC = alpha`
`rArr angle DAB = pi - (theta + alpha)`
`tan alpha = (p)/(q)`
In `DeltaABD`, using sine rule
`(AB)/(sin theta) = (BD)/(sin (pi - (theta - aplpha))) = (BD)/(sin (theta + alpha))`
`:. AB = (BD sin theta)/(sin (theta + alpha)) = (BD^(2) sin theta)/(BD sin (theta + alpha))`
`= (BD^(2) sin theta)/(BD sin theta cos alpha + BD cos theta sin alpha)`
`= ((p^(2) + q^(2)) sin theta)/(q sin theta + p cos theta)`