How to prove that the sum of the reciprocal of the squares of the two right sides of a right triangle is equal to the reciprocal of the high square on the bottom side

How to prove that the sum of the reciprocal of the squares of the two right sides of a right triangle is equal to the reciprocal of the high square on the bottom side

Using triangle area formula and Pythagorean theorem
S = AB / 2 = ch / 2, that is, ab = ch
a^2+b^2=c^2
1/a^2+b^2=(a^2+b^2)/a^2b^2=c^2/(ab)^2=c^2/(ch)^2=1/h^2

If the square of the hypotenuse of a right triangle is equal to 4 times the area of the right triangle, what are the two acute angles of the triangle?

2 ab = C * C = a * a + b * B
(a-b)*(a-b)=0
So a = B, so it's an isosceles right triangle. So it's all 45 degrees

If it is known that the square of the sum of the two right angles of the hypotenuse of a right triangle is equal to 6, then the area of the triangle is -

If the hypotenuse should be known, it can be calculated by the following method (it is not necessary to calculate the value of a and b). Let its two right angles be a and B respectively, and the hypotenuse be C, then a & # 178; + B & # 178; = C & # 178; if (a + b) &# 178; = 6, then 2Ab = 6 - (A & # 178; + B & # 178;) = 6-C & # 178;; ab = (6-C & # 178;) / 2