The incircle of an isosceles triangle $A B C, B A = B C$ , touches the sides $A B, B C$ and $C A$ at $D, E$ and $F$ respectively. Prove that $F$ bisects $A C$ .

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Solution

Given : $B A = B C$ ---(1)

$B D = B E$ (Tangents from point $B$ ) ---(2)

On subtracting equation (1) and (2), we get

$B A - B D = B C - B E$

$D A = E C$ ---(3)

$E C = C F$

$C F = A D$ ----(4)

$A D = A F$ ----(5)

Therefore, from equation (4) and (5),

$C F = A F$

Hence, $F$ bisects $A C$ .

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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.

Given : BA=BC ---(1)BD=BE (Tangents from point B) ---(2)On subtracting equation (1) and (2), we getBA−BD=BC−BEDA=EC ---(3)EC=CFCF=AD----(4)AD=AF----(5)Therefore, from equation (4) and (5),CF=AFHence, F bisects AC.

The incircle of an isosceles triangle $A B C, B A = B C$ , touches the sides $A B, B C$ and $C A$ at $D, E$ and $F$ respectively. Prove that $F$ bisects $A C$ .

Given : $B A = B C$ ---(1)

$B D = B E$ (Tangents from point $B$ ) ---(2)

On subtracting equation (1) and (2), we get

$B A - B D = B C - B E$

$D A = E C$ ---(3)

$E C = C F$

$C F = A D$ ----(4)

$A D = A F$ ----(5)

Therefore, from equation (4) and (5),

$C F = A F$

Hence, $F$ bisects $A C$ .