Hey please can anyone help me here, quickly the diagonals of the quadrilateral ABCD bisect each other perpendicularly at O.
prove that AB² + CD² = AD²+ BC²
no diagrams for the question.
i will give brainliest

In triangle ABD, we know, by pythagorean theorem

AD^2+BD^2=AB^2 - (I)

In triangle CBD, by pythagoreans theorem

CD^2+BD^2=BC^2 - (II)

Subtracting equation (I) and (II)

AD^2+BD^2-CD^2-BD^2=AB^2-BC^2

AD^2-CD^2=AB^2-BC^2

AD^2+BC^2=AB^2+CD^2

AB^2+CD^2=AD^2+BC^2

En triángulo ABD , lo sabemos, por Pitágoras teorema

ad² + bd² = ab²

picture below

step-by-step explanation:

x^2 -12x + 36 + y^2 = 36

x^2 + y^2 = 12x

r^2 = 12rcos(theta)

r = 12 cos(theta)