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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, $a+b=2\cdot \left(r+R\right)$.

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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 $CD = CE=\frac{a+b-c}{2}$, $AE = AF = \frac{b+c-a}{2}$, $BD = BF = \frac{a+c-b}{2}$.

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  • $\begingroup$ "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
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$$( a - r ) + (b - r ) = 2 R $$

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

$$\tag{3}(a+b)^2-(\sqrt{a^2+b^2})^2=2ab$$

which is true.

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