In triangle PQR, the side lengths are PQ=9, PR=10, and QR=17. Let X be the intersection of the angle bisector of angle P with side QR, and let Y be the foot of the perpendicular from X to side PR. Compute the length of XY.
+421 Hi guest, sorry for answering so late:
By the angle bisector theorem, QX / QP = RX / RP, and as we know RP and PQ, we can write QX / 9 = RX / 10. However, we also have that QX + RX = 17, so QX = 17 - RX. Thus, (17-RX)/9 = (RX)/10, and 170-10RX = 9RX, thus yielding RX = 170/19 and QX = 153 / 19.
Now, as PRX and QPX share an altitude from the same vertex, the ratios of their areas are the ratios of their bases. Thus, we can compute the area of PQR by heron's formula, which is 36. Thus, [PRX] / [PQX] = RX / QX --> 170/19 * 19/153 = 170/153 = 10/9. We also have that [PRX] + [PQX] = 36, so we can say that [PQX] = 36 - [PRX], so [PRX] / (36 - [PRX]) = 10/9, so 9[PRX] = 360-10[PRX], and [PRX] = 360/19.
Alternatively, we could have dropped an altitude and wrote equations to figure out the area of RPX.
We did all this to find XY. As XY is an altitude of PRX, we have (XY)(PR)/2 = 360/19, and as PR = 10, XY(5) = 360/19, so XY = 72/19
There are several different approaches to this problem, it's up to you to find which one works best!
