In triangle ABC, PQ drawn parallel to BC meets the side AB at P and the side AC at Q. The lines CP and BQ intersect each other at S. SR drawn parallel to AB meets BC at R. Prove that BR RC = AQ AC

Theorems: A line drawn parallel to a side of a triangle divides the other two sides proportionally. If two triangles are equiangular then the corresponding sides are proportional. If the three sides of a triangle are proportional to the three sides of another triangle, then the two triangles are equiangular.

This answer has been revised as a result of comments and corrections and is now believed to be correct. Point T has been inserted as the intersection of RS extended to meet PQ.

Similar triangles ABC and APQ: AQ/AC=PQ/BC.

To make it easier to see the similar triangles PQS and BCS, the points P', Q' and T' have been inserted such that P'Q' is the same length as PQ, making triangles PQS and P'Q'S congruent. It's then easier to appreciate the ratios of the sides of the similar triangles, because ST=S'T', PS=P'S, etc.

Similar triangles PQS and BCS: PQ/BC=QS/BS=PS/CS (angles QPS=BCS, PQS=CBS, PSQ=BSC, alternate angles between parallel lines). With the introduction of the extra points, we can also write P'Q'/BC=Q'S/BS=P'S/CS.

BR=PT in parallelogram BRTP and PT=P'T'. AQ/AC=PQ/BC=(PT+QT)/(BR+CR)=(BR+QT)/(BR+CR).

P'T'/CR=Q'T'/BR so PT/CR=QT/BR, because ST'/RS=ST/RS is a common ratio in triangles P'ST' and CRS, and Q'ST' and BRS.

But BR=PT so BR/CR=QT/BR, QT=BR^2/CR and AQ/AC=PQ/BC=(BR+BR^2/CR)/(BR+CR)=BR(CR+BR)/(CR(BR+CR))=BR/CR. Hence AQ/AC=BR/CR, in other words BR/RC=AQ/AC.

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Sorry. There's mistake in the question. Here is the right one.

In triangle ABC, PQ drawn parallel to BC meets the side AB at P and the side AC at Q. The lines CP and BQ intersect each other at S. SR drawn parallel to AB meets BC at R. Prove that BR/RC = AQ/AC

And Mr.Rod there is mistake in your picture. SR not  parallel to AC. It's parallel to AB.

Thank you for that and for pointing out my error. I'll start again.

I think this the best answer for this math. Thank you so much Mr.Rod.

Much appreciated, math93. Got there in the end!