Cords are intersecting and they are perpendicular to each other.
To prove: If two chords are intersecting and they are perpendicular to each other then four times the square of the radius is equal to the sum of squares of intercepts made by the chords.
Let intercepts are a, b, c, d.
\[4r^2 = a^2+b^2+c^2+d^2\]
Step 1:
Construction (Making a similar chord parallel to the original chord).
Step 2:
Construction (Making a right angled triangle).
Step 3:
It is a right angled triangle so that the line joining them is a diameter.
\[\Rightarrow (2r)^2=(c-a)^2+(b+d)^2\]
\[\Rightarrow 4r^2 = (a^2+c^2-2ac) + (b^2+d^2+2bd)\]
Similarly, we can do this with another chord and we will get:
\[\Rightarrow (2r)^2=(a+c)^2+(b-d)^2\]
\[\Rightarrow 4r^2 = (a^2+c^2+2ac) + (b^2+d^2-2bd)\]
Step 4:
By adding these two equations, we will get:
\[8r^2 = 2(a^2+c^2+b^2+d^2)\]
\[\Rightarrow 4r^2 = a^2+c^2+b^2+d^2\]
This is a standard form and can be used in many places.
One problem with the same concept:
