In triangle ABC, point M belongs to side BC. The segments AN and CM intersect at point P. Line BP crosses

| August 8, 2021 | education

In triangle ABC, point M belongs to side BC. The segments AN and CM intersect at point P. Line BP crosses the AC side at point R. What part of the AC is the segment AR if AM: MB = 3: 5 and NC: BN = 3: 2?

1. Draw through point B a straight line f parallel to the AC.

2. AN meets f at Q.

3. CM meets f at F.

4. The BNQ triangle is similar to the ANC triangle in two angles. Then BQ / AC = BN / NC = 2/3. Therefore, AC = 3/2 BQ.

5. The FBM triangle is similar to the AMC triangle in two angles. Then FB / AC = MB / AM = 5/3. Therefore, FB = 5/3 AC.

6. From (4) and (5) it follows that FB = 5/3 * 3/2 * BQ = 5/2 BQ. Therefore, BQ = 2/5 * FB = 2/5 * 5/3 * AC = 2/3 * AC.

7. The FPQ triangle is similar to the CPA triangle in two corners. Then QF / AC = FP / CB = PQ / PA.

QF = FB + BQ = 5/3 * AC + 2/3 * AC (From (5) and (6)) = 7/3 * AC.

QF / AC = 7/3 * AC / AC = 7/3 = FP / CP.

8. The FBP triangle is similar to the CRP triangle in two corners. Then FP / CP = FB / CR.

From (7) FB / CR = 7/3.

From (5) FB = 5/3 * AC.

Then CR = 3 * FB / 7 = 3 * 5/3 * AC / 7 = 5/7 * AC.

9. If CR = 5/7 * AC, then AR = 2/7 * AC, since AC = AR + CR.

Therefore, AC / AR = AC / (2/7 * AC) = 7/2, which is what was required to find.



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