For $\triangle ABC$ with right angle at $A$ and inradius $r$, show that $2r=b+c-a$

Question

I am trying to solve this problem:

The angle $A$ of $\triangle ABC$ is a right angle, and the sides of $BC$, $CA$ and $AB$ are of lengths $a$, $b$, and $c$, respectively. Each side of the triangle is tangent to a circle of radius r. Show that $2r=b+c-a$.

I have no idea on how to solve this problem, and could you guys help me with to solve it?

Thank you so much for you guys’ replies.

what do you mean by " Each side of the triangle to the circle S1 which is of radius r."? — user6767509, Jan 31 '20 at 12:26
Related (duplicate, when you realize the circumdiameter of a right triangle is its hypotenuse): "Radii of inscribed and circumscribed circles in right-angled triangle". See, in particular, my answer. — Blue, Jan 31 '20 at 12:32

score 1 · Accepted Answer · answered Jan 31 '20 at 15:17

1

Use $r=(S-a)tan(\frac{A}{2})$ $$ $$ Where $\angle A =90°$ hence $$r=s-a$$ $$2r=b+c-a$$

answered Jan 31 '20 at 15:17

Rajan

461

For $\triangle ABC$ with right angle at $A$ and inradius $r$, show that $2r=b+c-a$

1 Answers1