Question

A circle is inscribed in a $\Delta ABC$ having sides $8cm$, $10cm$, $12cm$ as shown in fig. find $AD,BF$ and $CE$

Answer 1

Let the center of the Circle be $O$

Then we can see that

$OD=OF$ (Radius)

$OB=OB$ (Common)

We can see that $△OBF$ is right angled as inscribed circle has sides of triangle perpendicular to it

$BF=\sqrt{O{B}^2-O{F}^2}$

Similarly, for $△OBD$

$BD=\sqrt{O{B}^2-O{D}^2}=\sqrt{O{B}^2-O{F}^2}$

Hence, $BF=BD=x\left(let\right)$

Similarly, $AD=AE=y\left(let\right)$

and $EC=CF=z\left(let\right)$

Now, we can see that

$BF+FC=12\Rightarrow x+z=\mathrm{12..........}\left(1\right)AE+EC=10\Rightarrow y+z=\mathrm{10...........}\left(2\right)BD+DA=8\Rightarrow x+y=\mathrm{8............}\left(3\right)$

Solving these equation simulataneously we will get

$x=5,y=3,z=7$

Now, AD=3

CE =7

BF=5