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

i don't have the answer rn but go to math papa. it shows the steps and the answer. hope this : )

step-by-step explanation:

hope it

step-by-step explanation:

x^(1/6) · x^(1/6) · x^(1/6)   ·x^(1/6) = x^(4/6)   since we add the exponents

4/6 can be simplified to 2/3

the simplified form is x^(2/3)