+569 In trapezoid ABCD with bases AB and CD, AB = 1, CD = 2, and ∠A = ∠D = 90◦ . P is a point on BC such that BP = 1 and PC = 2. Q is a point on DA such that PQ ⊥ BC. Find PQ.
+592 Because CPQ=QDC=90 and PC=CD=2, we know that PQ=QD. and for the exact same reasons,PQ=QA, or QD=QA. if we mark a point E bisecting CD, we can find AD by pythagorean theorum in triangle BCE:$\sqrt{9-1}=\sqrt{8}=2\sqrt2$, and AD/2=AQ=QD=PQ, so we have
PQ=$\frac{2\sqrt2}2=\boxed{\sqrt2}$
