Derékszögű háromszög-feladat, hogyan láthatnám be?

Köszönöm a gyors választ :)

A c=a+b-2r amúgy miért van így? Ezt eddig nem tudtam :) (csak kíváncsiságból :D)

Utánanéztem, és egy angol cikkben megtaláltam. Ez pofonegyszerű :D Belinkelem az eredetit, hátha valakit érdekel :) Köszönöm még1x a választ :)

owever, I would not solve the problem this way at all; there is a

much simpler formula that you can reach by simple geometry and algebra.

Draw the figure as follows: right triangle ABC with the right angle

at C, and a circle of radius r, with center O, inscribed in ABC. Call

the length AC = b and BC = a. Then draw perpendiculars from O to the

three sides, meeting AB, BC, and AC at D, E, and F respectively; and

join O to vertices A and B.

Now, triangles AOD and AOF are congruent, as are triangles BOD and

BOE. Also, OF is parallel to BC, and OE is parallel to AC. Thus CF =

CE = r, so AF = b-r and BE = a-r.

By the congruent triangles, AD = b-r and BD = a-r, so AB = a+b-2r.

But by the Pythagorean theorem, AB^2 = a^2 + b^2. Combining these results,

you can write a quadratic equation in r, with two solutions of which

the correct one is

r = (a+b-c)/2

where c = AB, the length of the hypotenuse.