What is the proof of the area of a triangle equals the exradius TIMES the semiperimeter minus the side in question?

Answer 1

The area of a triangle is the in-radius times the semi-perimeter. [This well-known result can be seen as follows: Connect the incenter to each vertex, thereby dividing the original triangle into three sub-triangles. Their areas are each half of base (a side) times height (the in-radius).]

Now, select a vertex, say A, and draw the ray through the incenter -- call it I. This ray also passes through the ex-center opposite A -- call it D. Drop perpendiculars from I and D to side AB (extended) -- call their feet E and F, respectively. Two similar right triangles are generated: AIE and ADF. Sides are proportional, so DF/IE = AF/AE. The left ratio is the ex-radius (opposite A) to the in-radius. The right ratio is the semi-perimeter to the semi-perimeter minus side BC (opposite A). Therefore ex-radius times (semi-perimeter minus BC) = in-radius times semi-perimeter, which is the triangle's area as shown above.

To see that AF is the semi-perimeter: Drop perpendiculars from D and I to AC (extended) -- call their feet G and H respectively. Let the excircle opposite A (which is centered at D) be tangent to BC at T. The two tangent segments from A to the excircle opposite it are equal: AG = AF. But likewise the two tangent segments from C are equal: CG = CT, and the two from B are equal: BF = BT. So the bent line ACT = ACG = ABF = bent line ABT, and since ACT and ABT between them cover the triangle's perimeter, all four are the semi-perimeter.

To see that AE is the semi-perimeter minus side opposite A: Drop a perpendicular from I to BC at J. Then the triangle's perimeter is divided into six pieces: BJ = BE (tangent segments), AE = AH (tangent segments), and CJ = CH (tangent segments). The semi-perimeter is the sum of one from each pair, e.g., AH + BJ + CJ. But BJ + CJ is the side opposite A, so AH alone is the semi-perimeter minus the side opposite A.

Rick Luttmann, Sonoma State University