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

the 3rd option is right

step-by-step explanation:

85.560%

step-by-step explanation:

conditional probability is defined as:

p(a|b) = p(a∩b) / p(b)

in other words, the probability that a occurs, given that b has occurred = the probability that both a and b occur / the probability that b occurs.

in this example:

p(turns 64 | turns 44) = p(turns 64 & turns 44) / p(turns 44).

if someone turns 64, they must have already turned 44, so:

p(turns 64 | turns 44) = p(turns 64) / p(turns 44)

on page 42, we see that of the original 100,000, 94944 turn 44, and 81234 turn 64.   therefore:

p(turns 64 | turns 44) = 81234 / 94944

p(turns 64 | turns 44) = 0.85560