Prove that

1)<APR = <AQR (< means angle)
2)AR and PQ bisect each other at right angles.

Please ignore the questions written in the attached file.
Just see the diagram and description of the diagram from there.

2. Hello, saha.subham!

Code:
              A
*
/|\
/ | \
/5 | 6\
P *---+---* Q
/ \3 | 4/ \
/   \ | /   \
/1    \|/    2\
B *-------*-------* C
R
$\displaystyle AB \,=\,AC$
$\displaystyle P,Q,R$ are midpoints of $\displaystyle AB, AC, BC$, respectively.

Prove that:

1) $\displaystyle \angle APR \,=\, \angle AQR$

2) $\displaystyle AR$ and $\displaystyle PQ$ bisect each other at right angles.

1) We are given: .$\displaystyle AB \,=\,AC.$
. . So: .$\displaystyle \Delta ABC$ is isosceles: .$\displaystyle \angle 1 \,=\, \angle 2$

Then: .$\displaystyle AP \,=\, PB \,=\, AQ \,=\, QC.$
Also: .$\displaystyle BR \,=\,RC$

Since $\displaystyle \Delta APQ$ is isosceles: .$\displaystyle \angle 5 \,=\,\angle 6$ .[1]

Note: .$\displaystyle \Delta PBR \,\cong\:\Delta QCR\;\;(s.a.s.)$
. And: .$\displaystyle PR \,=\,QR\;\text{ (corres. parts)}$
Then: .$\displaystyle \Delta PQR$ is isosceles: .$\displaystyle \angle 3 \,=\,\angle 4$ .[2]

Add [1] and [2]: .$\displaystyle \angle 5 + \angle 3 \;=\;\angle 6 + \angle 4$

Therefore: .$\displaystyle \angle APR \:=\:\angle 5 + \angle 3 \;=\;\angle 6 + \angle 4 \;=\;\angle AQR$

2) We have: .$\displaystyle \begin{array}{c}AP \,=\,AQ \\ PR \,=\,QR \end{array}$

$\displaystyle A$ is equidistant from the endpoints of segment $\displaystyle PQ.$
$\displaystyle R$ is equidistant from the endpoints of segment $\displaystyle PQ.$

Therefore, $\displaystyle AR$ is the perpendicular bisector of $\displaystyle PQ.$