Problem:
In â–³ABC,AB=8,BC=7,CA=6 and side BC is extended, as shown in the figure, to a point P so that â–³PAB is similar to â–³PCA. The length of PC is
Answer Choices:
A. 7
B. 8
C. 9
D. 10
E. 11
Solution:
By the similarity of the two triangles, PBPA​=PAPC​=ABCA​; hence PC+7PA​=PAPC​=86​, yielding the two equations
6(PC+7)=8PA and 6PA=8PC
From these one obtains PC=9.
