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

Question

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

Answer 1

Pranav,

 

Here's a more algebraic method:

 

Still, draw a diagram.  Triangle ABC with hypotenuse a and right angle A will be circumscribed by a circle whose diameter, D=a. 

 

The area ABC, Δ=bc/2

 

The diameter, d, of the inscribed circle is involved in the formula d=4Δ/p, where p is the perimeter.

See http://en.wikipedia.org/wiki/Incircle_and_excircles_of_a_triangle for a simple proof using areas.

 

2D+d=2a+2bc/p=2(ap+bc)/p=2(a^2+ab+ac+bc)/p  But a^2=b^2+c^2 by Pythagoras' Th.

 

So 2D+d=(a^2+b^2+c^2+2ab+2bc+2ac)/p=P^2/p=p

 

Answer 2

This is difficult to prove without providing a drawing, but here goes:

First, the hypotenuse of any circumscribed right triangle will have the length of the diameter of the circumcircle.  The center of the inscribed circle can be found by drawing the lines that bisect each angle of the triangle.  Picture a right triangle with a short leg = a, longer leg = b, and hypotenuse = c.  The angles opposite these sides will be A, B, and C.  Angle C is the right angle.  Draw the radius from the center of the inscribed circle to the point of tangency on side b, and the radius to the point of tangency on side a.  Both of these radii will form a right angle with the triangle side.  The two radii and pieces of side b from angle C to the point of tangency, and side a to the point of tangency will form a square having sides equal to the radius of the inscribed circle.  Now draw the radius from the inscribed circle center to the point of tangency on side c.  Again, a right angle is formed with the triangle.  The distance from angle A to the point of tangency on side c is equal to the distance from angle A to the point of tangency on side b.  The distance from angle A to the point of tangency on side b is also equal to the length b - the inscribed circle radius.  The distance from angle B to the tangency point on side c is equal to the distance from angle B to the tangency point on side a.  The distance from angle B to the tangency point on side a is also equal to a - the inscribed circle radius.  Now note that (a - the radius) + (b - the radius) = c.  The perimeter of the triangle is a + b + c.  This also equals 2c + 2 times the radius of the inscribed circle.  Therefore, the perimeter of the triangle is equal to twice the diameter of the circumcircle + the diameter of the inscribed circle.