Hello, shaurya!

I think I've got #3 . . .

Question 3

In Triangle ABC, the bisectors of angle B and angle C meet at P.

Through P, a straight line MPN is drawn parallel to BC.

Prove that MN = BM + CN Code:

A
*
*
* *
*
* *
*
* *
* *
* * P *
M *--------*---- N
* * * *
* * *
B *-----------------* C

Since BP is the bisector of angle B: $\displaystyle \angle PBM = \angle PBC$

Since $\displaystyle MN \parallel BC:\;\angle MPB = \angle PBC = \angle PBM$ (alt-int angles)

. . Hence: $\displaystyle \Delta MBP$ is isosceles and $\displaystyle BM = MP.$

Since CP is the bisector of angle C: $\displaystyle \angle PCN = \angle PCB$

Since $\displaystyle MN \parallel BC:\:\angle NPC = \angle PCB = \angle NCP$

. . Hence: $\displaystyle \Delta NPC$ is isosceles and $\displaystyle CN = PN.$

Then we have: .$\displaystyle MN \:=\:MP + PN \:=\:BM + CN$