The quadrilateral ABCD is a parallelogram with diagonal AC perpendicular to CD. The two diagonals intersect at E. How do you show that $DE^2 + 3EA^2 = AD^2$ ?

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The quadrilateral ABCD is a parallelogram with diagonal AC perpendicular to CD. The two diagonals intersect at E. How do you show that $DE^2 + 3EA^2 = AD^2$ ?

The question says to use Pythagoras' Theorem and to let CA=2d, CD=a and DA=c.

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Answer 1 · 2018-04-29T11:34:20

$"see explanation"$

Explanation:

$"using a "color(blue)"property of parallelograms"$

$• " the diagonals bisect each other"$

$rArrCE=EA" and "BE=DE$

$"using "color(blue)"Pythagoras "" in "triangleDCE$

$DE^2=CE^2+CD^2$

$rArrDE^2=EA^2+CD^2larr(CE=EA)$

$rArrCD^2=DE^2-EA^2$

$"using "color(blue)"Pythagoras ""in "triangleCDA$

$AD^2=CA^2+CD^2$

$color(white)(AD^2)=CA^2+DE^2-EA^2$

$color(white)(AD^2)=(2CE)^2+DE^2-EA^2$

$color(white)(AD^2)=4EA^2+DE^2-EA^2$

$color(white)(AD^2)=3EA^2+DE^2$

$rArrDE^2+3EA^2=AD^2larr" as required"$

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The quadrilateral ABCD is a parallelogram with diagonal AC perpendicular to CD. The two diagonals intersect at E. How do you show that $DE^2 + 3EA^2 = AD^2$ ?

The question says to use Pythagoras' Theorem and to let CA=2d, CD=a and DA=c.

1 Answer

Explanation:

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Math

Humanities

... and beyond

The quadrilateral ABCD is a parallelogram with diagonal AC perpendicular to CD. The two diagonals intersect at E. How do you show that DE^2 + 3EA^2 = AD^2DE^2 + 3EA^2 = AD^2?

The question says to use Pythagoras' Theorem and to let CA=2d, CD=a and DA=c.

1 Answer

Explanation:

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The quadrilateral ABCD is a parallelogram with diagonal AC perpendicular to CD. The two diagonals intersect at E. How do you show that $DE^2 + 3EA^2 = AD^2$ ?