+595 In triangle $PQR,$ let $X$ be the intersection of the angle bisector of $\angle P$ with side $\overline{QR}$, and let $Y$ be the foot of the perpendicular from $X$ to side $\overline{PR}$. If $PQ = 28,$ $QR = 35,$ and $PR = 17,$ then compute the length of $\overline{XY}$.
+129847 P
28 Y 17
Q X R
35
Because PX is a bisector, then
QX / PQ = RX / PR
PR / PQ = RX / QX
17/28 = RX / QX
So
RX = 17 / ( 17+ 28) * QR = (17 / 45) * 35 = 119/9
Law of Cosines
PQ^2 = QR^2 + PR^2 - 2(QR * PR) cos (PRQ)
28^2 = 35^2 + 17^2 - 2 (35 * 17) cos (PRQ)
cos (PRQ) = [ 28^2 - 35^2 - 17^2 ] / [-2 * 35 * 17 ] = 73/119
sin (PRQ) = sqrt [ 119^2 - 73^2 ] / 119 = 8sqrt (138) / 119
Law of Sines (angle XYR = 90 )
XY /sin (PRQ) = XR / sin(XYR)
XY / [8sqrt (138) / 119 ] = (119/9) / sin 90 (sin 90 = 1)
XY = [ 8sqrt (138) / 119 ] ( [ 119/ 9] = (8/9)sqrt (138) ≈ 10.44
