In △ABC the median AM is extended to ray AM and point P on AM is taken so that PM=AM. What is the distance from point P to vertices C and B if AB=c and AC=b?

Answer:PC = c and PB = bStep-by-step explanation:Given: As shown in figure 1 below:In ΔABC, AB = c and AC = b and AM is the median drawn from vertex A on BC and extended to point P e.g.,  PM = AM.To Find: PB =? and PC = ?Sol.  In ΔAMC and ΔBMPBM = MC ( ∵ Median from any vertex of a triangle divides its opposite side in two equal parts)∠BMP = ∠AMC (vertically opposite angles ∵ line BC and AP intersects each other at point M)PM  = AM (given)∴ ΔBMP ≅ ΔAMC (S-A-S postulate)∴ BP = AC = b ( corresponding parts of congruent triangles)Similarly, in  In ΔAMB and ΔCMPAM = MP ( ∵ Median from any vertex of a triangle divides its opposite side in two equal parts)∠AMB = ∠CMP (vertically opposite angles ∵ line BC and AP intersects each other at point M)BM  = MC (given)∴ ΔAMB ≅ ΔPMC (S-A-S postulate)∴ PC = AB = c ( corresponding parts of congruent triangles)