Question

In rectangle abcd, points p and q lie on side AB and DC respectively. Angle PMQ is a right angle, M is the midpoint of side BC and PB=4/3 BC. What is the ration PM: MQ

Answer 1

PM:MQ = 8:3.

Explanation:

`\rm \angle B\hat{M}P + 90^(\circ) + \angle C\hat{M}Q = 180^(\circ)`;

`\implies \rm \angle B\hat{M}P + \angle C\hat{M}Q = 90^(\circ)`;

`\implies \rm 90^(\circ) - \angle B\hat{M}P = \angle C\hat{M}Q`.

Also,

`\rm \angle B\hat{P}M = 90^(\circ) - \angle B\hat{M}P` in right triangle PBM.

Thus
`\rm \angle{P\hat{B}M} = \angle C\hat{M}Q`.

Additionally
`\rm \angle \hat{B} = 90^(\circ) = \angle \hat{C}`.

Therefore
`\rm \triangle PBM \sim \triangle MCQ`.

`\rm \displaystyle BC = 2\;MC` for M is the midpoint of segment BC.

`\rm \displaystyle PB = (4)/(3)BC = (8)/(3)MC`.

`\rm \triangle PBM \sim \triangle MCQ` implies that

`\displaystyle \rm PM:MQ = PB:MC = 1:(8)/(3) = 8:3`.