Question
Show that the angle bisectors of a triangle are concurrent
✓
Answer
Given: ABC is a triangle. AD, BE and CF are the angle bisector of ∠A, ∠B, and ∠C.
To Prove: Bisector AD, BE and CF intersect
Proof: The angle bisectors AD and BE meet at O.
Assume CF does not pass through O. By angle bisector theorem.
AD is the angle bisector of ∠A
$\frac{ BD }{ DC }=\frac{ AB }{ AC } \ldots(1)$
$BE$ is the angle bisector of $\angle B$
$
\frac{C E}{E A}=\frac{B C}{A B}
$
$C F$ is the angle bisector $\angle C$
$
\frac{ AF }{ FB }=\frac{ AC }{ BC } \ldots(3)
$
Multiply (1) (2) and (3)
$
\frac{ BD }{ DC } \times \frac{ CE }{ EA } \times \frac{ AF }{ FB }=\frac{ AB }{ AC } \times \frac{ BC }{ AB } \times \frac{ AC }{ BC }
$
So by Ceva's theorem.
The bisector $A D, B E$ and $C F$ are concurrent.
To Prove: Bisector AD, BE and CF intersect
Proof: The angle bisectors AD and BE meet at O.
Assume CF does not pass through O. By angle bisector theorem.
AD is the angle bisector of ∠A
$\frac{ BD }{ DC }=\frac{ AB }{ AC } \ldots(1)$
$BE$ is the angle bisector of $\angle B$
$
\frac{C E}{E A}=\frac{B C}{A B}
$
$C F$ is the angle bisector $\angle C$
$
\frac{ AF }{ FB }=\frac{ AC }{ BC } \ldots(3)
$
Multiply (1) (2) and (3)
$
\frac{ BD }{ DC } \times \frac{ CE }{ EA } \times \frac{ AF }{ FB }=\frac{ AB }{ AC } \times \frac{ BC }{ AB } \times \frac{ AC }{ BC }
$
So by Ceva's theorem.
The bisector $A D, B E$ and $C F$ are concurrent.
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