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
$AB \,=\,AC$
$P,Q,R$ are midpoints of $AB, AC, BC$, respectively.

Prove that:

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

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

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

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

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

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

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

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

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

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

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