Two externally tangent circle

Given : M is the midpoint of BC and as shown in figure .
Prove: BC = 2 AF

Solution

Given : M is the midpoint of BC and as shown in figure .
Prove: BC = 2 AF
$ \displaystyle \text{Proof}\ :\ \Delta \ BHA\sim \Delta \ BGC\ \ \left( {\because HA\parallel GC} \right)$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \frac{{AB}}{{AH}}=\frac{{BC}}{{CG}}\ \ .......(1)$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \text{In}\ \Delta \ BHA\ \ \text{and}\ \Delta \ FMA\ ,$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \angle \ BHA=\angle \ FMA={{90}^{\circ }}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \angle \ BAH=\angle \ FAM\ \ \left( {\because \text{opposite}\ \angle \text{s}} \right)$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \Delta \ BHA\sim \Delta \ FMA\ \ \left( {AA\ \text{Cor}:} \right)$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \frac{{AB}}{{AH}}=\frac{{AF}}{{AM}}\ \ .......(2)$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \text{From}\ (1)\ \text{and}\ (2),$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \frac{{BC}}{{CG}}=\frac{{AF}}{{AM}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \frac{{BC}}{{R-r}}=\frac{{AF}}{{\frac{1}{2}(R+r)-r}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \frac{{BC}}{{R-r}}=\frac{{2AF}}{{R-r}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \therefore \ BC=2AF$