\( ABC, AP = \frac{PQ}{2} = BQ$ $\frac{CR}{RA} = \frac{3}{2}. \frac{[BQC]}{[CRQ]}.$\) in triangle ABC, AP = PQ/2 = BQ and CR/RA=3/2. Find the area of BQC/(the area of)CRQ.
+26367 In triangle ABC, AP = PQ/2 = BQ and CR/RA=3/2.
Find the area of BQC/(the area of)CRQ.
In triangle
\(\text{In triangle } ABC,\ AP = \frac{PQ}{2} = BQ,\quad\dfrac{CR}{RA} = \dfrac{3}{2}. \dfrac{[BQC]}{[CRQ]}. \)
My answer see here: https://web2.0calc.com/questions/geometry-problem-im-stuck-on#r3
