Exercise 1261Geometry › Plane geometryL1Added 4 Nov 2026

The area of a right triangle from the incircle touch point

#incircle#tangent lengths#Pythagoras#right triangle

Statement

Let $ABC$ be a triangle with a right angle at $C$ , and let its incircle touch the hypotenuse $AB$ at $T$ . Let $x=AT$ and $y=TB$ . Prove that the area of $ABC$ equals $xy$ . Check it on the triangle with legs $5$ and $12$ .

Hint

Let $r$ be the inradius. The quadrilateral formed with $C$ and the two touch points on the legs is a square of side $r$ . Use tangent lengths and Pythagoras.

Show solutionHide solution

Solution

Tangent lengths. The legs have lengths $CA=x+r$ and $CB=y+r$ . The hypotenuse is $AB=x+y$ .

Pythagoras.
$(x+r)^2+(y+r)^2=(x+y)^2\iff r^2+r(x+y)=xy.$

Area. The semi-perimeter is $s=x+y+r$ . The area is
$[ABC]=rs=r(x+y+r)=r^2+r(x+y)=xy.\ \blacksquare$

Check. Legs $5,12$ , hypotenuse $13$ : $s=15$ , $r=2$ . With $CA=12$ , $CB=5$ : $x=10$ , $y=3$ , $xy=30=\frac{5\cdot12}2$ ✓.

Remark. As a by-product, $r=\frac{CA+CB-AB}2$ .

Lesson. Describe all lengths in terms of tangent lengths and the inradius; then Pythagoras becomes an algebraic identity.

The area of a right triangle from the incircle touch point

Statement

Solution

More problems in Plane geometry