A simple "Geometry Revisited" problem...

[photon]

let p radii circle $P$ and q radii circle $Q$.$B$ lies in $P$ and $C$ lies in $Q$.
take any point $B'$ from $P$ circumference and $C'$ from $Q$ as well.add $B'$ to $B$ and $A$ and add $C'$ to $C$ and $A$.
for ekantor brittangshostho cone.
\[\angle B'=\angle B,\angle C'=\angle C\]
now,\[\frac{AB}{sinB'}=2p.........(1)\]
\[\frac{AC}{sinC'}=2q.......(2)\]
in triangle ABC,
\[\frac{AC}{sinB}=2R........(3)\]
\[\frac{AB}{sinC}=2R..........(4)\]
multiplying 1 and 2,\[\frac{AB.AC}{sinB'sinC'}=4pq\]
\[or,\frac{AB.AC}{sinBsinC}=4pq[\because sinB=sinB',sinC=sinC']\]
multyplying 3 and 4 we get,\[\frac{AB.AC}{sinBsinC}=4R^{2}\]
then,$4R^2$=4pq$ $R^2=pq$

But I ain't Sherlock Homes. I can't stand mysteries.

won't trouble you more....

it's samiur,dhaka.

[Tahmid Hasan]

just applying sine law and alternate segment tpgeorem would be enough