Exercise 1259Geometry › Plane geometryHigh schoolAdded 4 Nov 2026

The Gergonne point and Ceva's theorem

#Ceva's theorem#incircle#tangent lengths

Statement

The incircle of triangle $ABC$ touches $BC,CA,AB$ at $D,E,F$ respectively.

Prove that $AD$ , $BE$ , $CF$ are concurrent.
For the triangle with $BC=14$ , $CA=15$ , $AB=13$ , find the ratio $BD:DC$ .

Hint

You may use Ceva's theorem (converse): if $D,E,F$ are on the sides $BC,CA,AB$ and $\frac{AF}{FB}\cdot\frac{BD}{DC}\cdot\frac{CE}{EA}=1$ , then $AD,BE,CF$ are concurrent. Two tangent segments from the same point to a circle are equal.

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Solution

1. Tangent lengths. Let $x=AE=AF$ , $y=BF=BD$ , $z=CD=CE$ . Then
$\frac{AF}{FB}\cdot\frac{BD}{DC}\cdot\frac{CE}{EA}=\frac xy\cdot\frac yz\cdot\frac zx=1,$
and the points $D,E,F$ are interior to the sides. By Ceva's converse, $AD,BE,CF$ concur (at the Gergonne point). $\blacksquare$

2. The numbers. Adding, $x+y=AB=13$ , $y+z=BC=14$ , $z+x=CA=15$ . The semi-perimeter is $s=21$ and $x=s-a=7$ , $y=s-b=6$ , $z=s-c=8$ , where $a=BC,b=CA,c=AB$ . Then
$BD:DC=y:z=6:8=3:4.$

Remark. In general $AE=AF=s-a$ , $BD=BF=s-b$ , $CD=CE=s-c$ .

Lesson. Name the three tangent lengths $x,y,z$ ; most incircle problems reduce to linear equations in them.

The Gergonne point and Ceva's theorem

Statement

Solution

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