Question

Prove that in a right-angled triangle the sum of diameter of incircle and diameter of circumcircle is equal to sum of the sides containing the right angle.

Accepted Answer

Let $∆ A B C$ be a right-angled triangle.

And the circles with centres $M and N$ are incircle and circumcircle.

Let $M D, M E and M F$ are radii of the incircle.
So, $M D = M E = M F = r_{1}$
Also, $M D ⊥ B C and M F ⊥ A B$ .

So, $B F = B D = r_{1}$ (by the property of tangents to the circle from an external point)
Now $N A and N C$ are radii of the circumcircle.
$N A = N C = r_{2}$
So, diameter of circumcircle $= A C = 2 r_{2}$ -----(i)

Now, the sides of a triangle are tangents of the incircle.
$A B and A C$ are tangents to the incircle from point $A$ , meeting the incircle at points $F and E$ , respectively.
So, $A F = A E$ (by the property of tangents to a circle from an external point)
Similarly, $C B and C A$ are tangents to the incircle from point $C$ , meeting the incircle at
points $D and E$ , respectively.
So, $C D = C E$ (by the property of tangents to a circle from an external point)
So, $A F + B F = A E + r_{1}$ -----(ii) and $B D + D C = r_{1} + E C$ -----(iii)
Now, from equation (ii) and (iii),

$A B + B C = A F + B F + B D + D C$

$\Rightarrow A B + B C = A E + r_{1} + r_{1} + E C$

$\Rightarrow A B + B C = 2 r_{1} + A C$

$\Rightarrow A B + B C = 2 r_{1} + 2 r_{2}$ [from equation (i)]

Where, $r_{1} and r_{2}$ are radii of incircle and circumcircle respectively.
So, in a right-angled triangle the sum of diameter of incircle and diameter of circumcircle is equal to sum of the sides containing the right angle.

Hence, proved.

Prove that in a right-angled triangle the sum of diameter of incircle and diameter of circumcircle is equal to sum of the sides containing the right angle.

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