Question

1. ABC is an acute-angled triangle. P & Q are the points on AB and AC respectively and area of △APC = area of △AQB. A line is drawn through B parallel to AC and meets the line through Q parallel to AB at S. QS cuts BC at R. Prove that RS = AP ​

Answer 1

In quadrilateral ABCD we have                  AC = AD            and AB being the bisector of ∠A.            Now, in ΔABC and ΔABD,                  AC = AD[Given]                  AB = AB[Common]∠CAB = ∠DAB [∴ AB bisects ∠CAD]            ∴ Using SAS criteria, we have                  ΔABC ≌ ΔABD.            ∴ Corresponding parts of congruent triangles (c.p.c.t) are equal.            ∴ BC = BD.