Prove that the perimeter of a right triangle is equal to the sum of the diameter of its incircle and twice the diameter of its circumcircle.
Prove that the perimeter of a right triangle is equal to the sum of the diameter of its incircle and twice the diameter of its circumcircle.

Selina Solutions Concise Class 10 Maths Chapter 18 ex. 18(C) - 13

Join

    \[OL,\text{ }OM\text{ }and\text{ }ON\]

Suppose,

    \[D\text{ }and\text{ }d\]

be the diameter of the circumcircle and incircle.

Also,

    \[R\text{ }and\text{ }r\]

be the radius of the circumcircle and incircle.

Now, in circumcircle of 

    \[\vartriangle ABC,\]

    \[\angle B\text{ }=\text{ }{{90}^{o}}\]

Thus,

    \[AC\]

is the diameter of the circumcircle

    \[i.e.\text{ }AC\text{ }=\text{ }D\]

Let the radius of the incircle be

    \[r\]

    \[OL\text{ }=\text{ }OM\text{ }=\text{ }ON\text{ }=\text{ }r\]

From

    \[B,\text{ }BL\text{ }and\text{ }BM\]

are the tangents to the incircle.

So,

    \[BL\text{ }=\text{ }BM\text{ }=\text{ }r\]

Similarly,

    \[AM\text{ }=\text{ }AN\]

And

    \[CL\text{ }=\text{ }CN\text{ }=\text{ }R\]

[Tangents from the point outside the circle]

Now,

    \[AB\text{ }+\text{ }BC\text{ }+\text{ }CA\]

    \[=\text{ }AM\text{ }+\text{ }BM\text{ }+\text{ }BL\text{ }+\text{ }CL\text{ }+\text{ }CA\]

Or,

    \[=\text{ }AN\text{ }+\text{ }r\text{ }+\text{ }r\text{ }+\text{ }CN\text{ }+\text{ }CA\]

    \[=\text{ }AN\text{ }+\text{ }CN\text{ }+\text{ }2r\text{ }+\text{ }CA\]

Or,

    \[=\text{ }AC\text{ }+\text{ }AC\text{ }+\text{ }2r\]

    \[=\text{ }2AC\text{ }+\text{ }2r\]

So,

    \[=\text{ }2D\text{ }+\text{ }d\]

– Hence Proved