Question

if two times the square of diameter of the circumcircle of a triangle is equal to the sum of the square of its side then prove that the triangle is right angle

Answer 1

For a better understanding of the solution provided here please find the diagram in the file attached. Also, extremely important in this case is the following fact:

"In the case of a right triangle, the hypotenuse is a diameter of the circumcircle, and its center is exactly at the midpoint of the hypotenuse."

Let the radius of the circle be r. Therefore, it's diameter will be 2r. Let the sides be a,b,c and let M be the center of the circle. Then by the above statement, RM=MQ=r.

Thus, we will have the length c to be as: c=2r. Therefore, if we can prove that c=2r, we will prove that triangle PQR is a right triangle.

Now, let us proceed with what has been given to us.

$2\times(2r)^2=a^2+b^2+c^2$

$8r^2=a^2+b^2+c^2$..................Equation 1

Now, if PQR is the right triangle then, from the diagram:

$PQ^2+PR^2=QR^2$

$a^2+b^2=c^2$............Equation 2

Now, plugging in Equation 2 into Equation 1 we get:

$c^2+c^2=8r^2$

$2c^2=8r^2$

$\therefore c=2r$

Since we have successfully proved that c=2r, we have proved that PQR is a right triangle.