If three line segments a, B and C satisfy a square = C square - b square, is the triangle formed by these three line segments a right triangle? Why? Please explain the process and reasons,

If three line segments a, B and C satisfy a square = C square - b square, is the triangle formed by these three line segments a right triangle? Why? Please explain the process and reasons,

a²=c²-b²'
So a & # 178; + B & # 178; = C & # 178;
So it's a right triangle
This is
Pythagorean theorem
That is, if the sum of squares of two sides of a triangle is equal to the square of the other side
Then the triangle is a right triangle

If three line segments A.B.C satisfy that the square of a = the square of C - the square of B, is the triangle formed by these three line segments a right triangle? Why?

From the cosine theorem C & # 178; = A & # 178; + B & # 178; - 2abcosc
Then COSC = (A & # 178; + B & # 178; - C & # 178;) / (2Ab)
It is known that a & # 178; = C & # 178; - B & # 178;
Then a & # 178; + B & # 178; - C & # 178; = 0
That is, COSC = 0
Then C = 90 degree
So it is a right triangle

If the two right sides of a right triangle are a and B respectively, and the height of the hypotenuse is h, then A2 / 1 + B2 / 1 =?

It is known that a, B and C are the lengths of the three sides of the right triangle ABC, and 2 (A2 + B2) 2 - (A2 + B2) = 6

Let A2 + B2 = X
Then 2x2-x-6 = 0
Change to (2x + 3) (X-2) = 0
Then A2 + B2 = x = 2
So C = root 2