Exercise 1316Geometry › Plane geometryOlympiadAdded 5 Nov 2026

Excircles and tangent lengths

#excircle#incircle#tangent lengths#semiperimeter#area

Statement

Let $ABC$ have sides $a=BC$ , $b=CA$ , $c=AB$ , semiperimeter $s=\frac{a+b+c}2$ , area $S$ , inradius $r$ . The incircle touches $BC$ at $D'$ . The excircle opposite $A$ has radius $r_a$ and touches $BC$ at $D$ , line $AB$ at $E$ and line $AC$ at $F$ .

Prove $AE=AF=s$ .
Prove $BD=s-c$ and $CD=s-b$ , and $BD'=s-b$ , $CD'=s-c$ . Conclude that $D$ and $D'$ are symmetric with respect to the midpoint of $BC$ .
Prove $S=r_a(s-a)$ , and deduce $r\,r_a=(s-b)(s-c)$ .

Hint

Two tangent segments from the same point to a circle are equal. For part 3 split the area of $ABC$ using the excentre $I_a$ .

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Solution

1. Tangent segments from $B$ give $BE=BD$ and from $C$ give $CF=CD$ ; from $A$ , $AE=AF$ . Since $E$ lies on the extension of $AB$ beyond $B$ and $F$ on the extension of $AC$ beyond $C$ ,
$AE+AF=(AB+BE)+(AC+CF)=c+b+(BD+DC)=a+b+c=2s.$
Hence $AE=AF=s$ .

2. $BD=BE=AE-AB=s-c$ and $CD=CF=AF-AC=s-b$ . For the incircle, let $x=AE'=AF'$ , $y=BD'=BF'$ , $z=CD'=CE'$ . Then $x+y=c$ , $x+z=b$ , $y+z=a$ , so $y=s-b$ and $z=s-c$ . Therefore $BD=s-c=CD'$ and $CD=s-b=BD'$ .

3. Let $I_a$ be the excentre. Then
$[ABC]=[ABI_a]+[ACI_a]-[BCI_a]=\tfrac12cr_a+\tfrac12br_a-\tfrac12ar_a=r_a(s-a).$
Also $S=rs$ , and Heron's formula gives $S^2=s(s-a)(s-b)(s-c)$ . Multiplying $S=rs$ and $S=r_a(s-a)$ :
$S^2=r\,r_a\,s(s-a)\ \Longrightarrow\ r\,r_a=(s-b)(s-c).$

Remark. The distance between the two touch points on $BC$ is $|b-c|$ .

Lesson. Tangent-length bookkeeping turns circle configurations into linear equations in the semiperimeter.

Excircles and tangent lengths

Statement

Solution

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