# Prove $\frac{1}{r} = \frac{1}{a} + \frac{1}{b}$ for a semicircle tangent within a right triangle

I have a right angled triangle with the sides which are not hypotenuses $$a$$ and $$b$$. There is also a semi-circle radius $$r$$ whose diameter lies on the hypotenuse of the right angles triangle, and sides $$a$$ and $$b$$ are tangents of the semicircle.

Prove that $$\frac{1}{r} = \frac{1}{a} + \frac{1}{b}$$

My attempt: First I drew the diagram:

I marked all areas of significance:

The vertices opposite side $$a$$ is $$A$$ , $$b$$ to $$B$$ and $$r$$ to $$C$$. Where the semi circle touches $$a$$, I called $$P$$, and where it touched $$b$$ I called $$Q$$. The centre of the semicircle I call $$O$$.

But then, I couldn't proceed. What do I do after this?

We can write the area of $\triangle ABC$ in two ways: $$Area = \frac{1}{2}\cdot ab$$ $$Area = \frac{1}{2}\cdot a\cdot OP+\frac{1}{2}\cdot b\cdot OQ = \frac{1}{2}\cdot (ar+br)$$ Equate these two equations and divide both sides by $\frac{1}{2}\cdot abr$ to get your result: $$\frac{1}{2}\cdot ab=\frac{1}{2}\cdot (ar+br) \implies \frac{1}{r}=\frac{1}{a}+\frac{1}{b}$$

• Ooh, you beat me to it. Deleting. May 17, 2017 at 15:49
• @BrianTung, Please don't delete it ; it really doesn't matter. May 17, 2017 at 15:55
• @Goodra: Thanks, but it really is redundant. Deletion decreases clutter, improving readability, and I don't need reputation that much. :-) May 17, 2017 at 17:00

Note :

$OP || AB$ (side c);

$OQ || AC$ (side a).

Similar triangles:

$\triangle BOQ$ is similar to $\triangle BCA$.

$r/a$ = $(length BQ) / b$.

$(length BQ) = b - r$, since

$AQOP$ is a square, side length $r$.

Combining:

$r/a = (b-r)/b$, or $r/a = 1 - r/b$,

dividing both sides by $r$:.

$1/r = 1/a +1/b$.