Answer:
PC = c and PB = b
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 ΔBMP
BM = 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 ΔCMP
AM = 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)