Problem:
In isosceles trapezoid ABCD, parallel bases AB and CD have lengths 500 and 650 , respectively, and AD=BC=333. The angle bisectors of ∠A and ∠D meet at P, and the angle bisectors of ∠B and ∠C meet at Q. Find PQ.
Solution:
Answer (242):
Let point K lie on BC so that QK∥AB. Because ∠KBQ=∠QBA=∠BQK, it follows that △KBQ is isosceles with BK=QK. Similarly, △KCQ is isosceles with CK=QK. Hence BK=QK=CK. Analogously, let point J lie on AD so that PJ∥AB. Then J is the midpoint of AD, and AJ=PJ=DJ. Thus P and Q lie on line KJ, which is the midline of the trapezoid with KJ∥AB, and KJ=2AB+CD=575.
Then
PQ=KJ−KQ−PJ=KJ−2⋅PJ=KJ−AD=575−333=242.
OR
Let E be the intersection of lines AB and DP, and let F be the intersection of lines CD and AP. Let G,H, and I be the projections of points A,P, and Q onto line CD, respectively.
Because angles ∠BAD and ∠CDA are supplementary, it follows that angles ∠PAD and ∠PDA are complementary and AF⊥DE. Triangle △FDA is isosceles because PD is both an altitude and an angle bisector of △FDA. Similarly, △EAD is isosceles, and thus AE=AD=DF=333. Because ABCD is an isosceles trapezoid, DG=21(CD−AB)=75. Because P is the midpoint of AF,H is the midpoint of GF, and thus
GH=21⋅GF=21(DF−DG)=21(333−75)=129
and DH=DG+GH=204. By symmetry, CI=DH=204. The requested length is therefore
