Question

The incircle of an isosceles triangle $ABC,BA=BC$, touches the sides $AB,BC$ and $CA$ at $D,E$ and $F$ respectively. Prove that $F$ bisects $AC$.

Answer 1

Hint: Here first we have to draw the figure. circles are equal. We are given that $BA=BC$ where $BA=BD+AD$ and $BC=BE+EC$. Now, by applying the theorem that the lengths of the tangents drawn from an external point to the circle are equal we will get $AF=FC$.

Complete step by step answer:
First, let us draw the figure.

Given that $\Delta ABC$ is an isosceles triangle where $BA=BC$ and the triangle touches the circle at $D,E$ and $F$ respectively.
We have to prove that F bisects $AC$ i.e.$AF=FC$
We have $BA=BC$.
We know by the theorem that the lengths of the tangents drawn from an external point to a circle are equal. Hence we can say write:
$\begin{align}
  & AD=AF \\
 & EC=FC \\
\end{align}$
From the figure we can say that:
$BA=BD+AD$
We have already got, $AD=AF$. Hence we can write:
$BA=BD+AF$ ….. (1)
From the figure we can also say that:
$BC=BE+EC$
Since, $EC=FC$, we can write:
$BC=BE+FC$ ….. (2)
We are given that $BA=BC$, therefore, from equation (1) and equation (2) we can write:
$BD+AF=BE+FC$ …. (3)
Again by the theorem that the lengths of the tangents drawn from an external point to the circle are equal, we can write:
$BD=BE$
By applying this to equation (3) we obtain:
$BE+AF=BE+FC$
Now by taking $BE$ to the left side, $BE$ becomes $-BE$. Hence we will get:
$BE+AF-BE=FC$
Next, by cancellation of $BE$ and $-BE$ we obtain:
$AF=FC$
Therefore, we can say that \[F\] is the midpoint of \[AC\].
Hence, \[F\] bisects \[AC\].

Note: Here, you can also prove this by congruence of triangles. For that, consider the centre of the circle to be \[O\]. Join \[OA,\text{ }OC\]and \[OF\]. Then, you will get two triangles $\Delta OFA$ and $\Delta OFC$. Now, you have to show that $\Delta OFA\cong \Delta OFC$ by ASA rule. Then, you will get $AF=FC$.