Question

A circle touches all the three sides of a right angled AABC in which ∠B is right angle. Prove that the radius of the circle is AB+BC–AC/2.

Answer 1

A circle touches all the three sides of a right angled triangle ∆ABC in which B is right angle.

we have to prove that radius of circle
$\bf{r=\frac{AB+BC-AC}{2}}$

proof :- Let O is the centre of circle and radius of circle is r. Let D , E , F are the points where OD ⊥ BC , OE ⊥ CA and OF ⊥ AB.
see figure,
area of ∆ABC = area of ∆AOB + area of ∆BOC + area of ∆COA
1/2 × AB × BC = 1/2 × OD × BC + 1/2 × OE × CA + 1/2 × OF × AB

as you can see
OE = OF = OD = radius of incircle = r

so, AB × BC = r × BC + r × CA + r × AB
⇒ AB × BC = r(AB + BC + CA) .......(i)

∆ABC is right angled triangle,
so, AC² = AB² + BC²
⇒AC² = (AB + BC)² - 2AB × BC
⇒2AB × BC = (AB + BC)² - AC²
⇒ 2AB × BC = (AB + BC + AC)(AB + BC - AC).....(ii)

from equations. (i) and (ii),
(AB + BC + AC)(AB + BC - AC)/2 = r(AB + BC + CA)
r = (AB + BC - AC)/2
hence proved.

Answer 2

Answer:

good.............m...............