In a right angled triangle, △ ABC, with sides a and b adjacent to the right angle, the radius of the inscribed circle is equal to r and the radius of the circumscribed circle is equal to R.

Prove that in △ABC, $a+b=2\cdot \left(r+R\right)$.


Another method is to notice that, by letting $D,E,F$ be the points where the inscribed circle meets $BC, AC, AB$, we know that $r = CD = CE$ (because CD,CE and two radii form a square), hence $$r = CD = \frac{a+b-c}{2} \implies a+b = c+2r = 2(r+R)$$

Note: The length of CD comes from solving the linear equation $BD + CD = a$, $AE + CE = b$, $AF + BF = c$, and $CD = CE$, $AE = AF$, $BD = BF$ to get $CD = CE=\frac{a+b-c}{2}$, $AE = AF = \frac{b+c-a}{2}$, $BD = BF = \frac{a+c-b}{2}$.

  • $\begingroup$ "Easier" method ? Say "another" method... $\endgroup$ – Jean Marie Apr 23 '17 at 15:41
  • $\begingroup$ Fixed :) thank you for pointing it out! $\endgroup$ – Lazy Lee Apr 24 '17 at 0:16

enter image description here

$$( a - r ) + (b - r ) = 2 R $$

  • 2
    $\begingroup$ [+1] Nice "proof without words" $\endgroup$ – Jean Marie Apr 23 '17 at 15:44

We have the following two formulas for $R$ and $r$:

  • $$R=\tfrac12c=\tfrac12 \sqrt{a^2+b^2}$$

($c^2=a^2+b^2$ by Pythagoras.)

  • $$r=\dfrac{ab}{a+b+c}$$

(proof: the area of the (right) triangle ABC, i.e., $\tfrac12ab$ is equal to the sum of the areas of triangles $IBC, ICA, IAB$, i.e., $\tfrac12ra+\tfrac12rb+\tfrac12rc$, where $I$ is the center of the inscribed circle.)

Thus we have to prove that

$$\tag{1}a+b=2\dfrac{ab}{a+b+\sqrt{a^2+b^2}}+2\tfrac12 \sqrt{a^2+b^2}$$

In fact, (1) is equivalent with:

$$\tag{2}a+b-\sqrt{a^2+b^2}=\dfrac{2ab}{a+b+\sqrt{a^2+b^2}} $$

itself equivalent to:


which is true.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.