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:rightang and circ

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?


2 Answers 2


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: $$\frac{1}{2}\cdot ab=\frac{1}{2}\cdot (ar+br) \implies \frac{1}{r}=\frac{1}{a}+\frac{1}{b}$$

  • 1
    $\begingroup$ Ooh, you beat me to it. Deleting. $\endgroup$
    – Brian Tung
    May 17, 2017 at 15:49
  • 1
    $\begingroup$ @BrianTung, Please don't delete it ; it really doesn't matter. $\endgroup$
    – Xetrov
    May 17, 2017 at 15:55
  • $\begingroup$ @Goodra: Thanks, but it really is redundant. Deletion decreases clutter, improving readability, and I don't need reputation that much. :-) $\endgroup$
    – Brian Tung
    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 $r$.


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

dividing both sides by $r$:.

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


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