Problem: In triangle ABC lines CE and AD are drawn so that DBCD​=13​ and EBAE​=23​. Let r=PECP​ where P is the intersection point of CE and AD. Then r equals:
Answer Choices:
A. 3
B. 23​
C. 4
D. 5
E. 23​
Solution:
Draw \mathrm{DR} \| \mathrm{AB} \cdot \dfrac{\overline{\mathrm{CR}}}{\overline{\mathrm{RE}}}=\dfrac{\overline{\mathrm{CD}}}{\overline{\mathrm{DB}}}=\dfrac{3}{1}, \dfrac{\overline{\mathrm{RD}}}{\overline{\mathrm{EB}}}=\dfrac{\overline{\mathrm{CD}}}{\overline{\mathrm{CB}}}=\dfrac{3}
∴CR=3RE=3RP+3PE and RD=43​ EB
∴CP=CR+RP=4RP+3PE.
Since △RDP∼△EAP,PERP​=AERD​
∴RD=PERP×AE​. But AE=23​ EB,∴RD=PERP​⋅23​ EB
∴43​ EB=23​ EB⋅PERP​,RP=21​PE,CP=4⋅21​PE+3PE=5PE∴PECP​=5
