Geometry question (1 Viewer)

[acmilan]

The quadrilateral ABCD is a parallelogram with diagonal AC perpendicular to DC. The two diagonals intersect at E.

Show that DE^2 + 3EA^2 = AD^2

 

[smallcattle]

CD^2 + CE^2 = DE^2

CD^2 + (CE+EA)^2 = DA^2

CD^2 + CE^2 + 2CEEA + EA^2 = DA^2

DE^2 + 2CEEA + EA^2 = DA^2

CE=EA (diagonals bisect each other in parallelgram)

DE^2 + 2EA^2 + EA^2 = DA^2

therefore DE^2 + 3EA^2 = DA^2

 

[JayWalker]

CD^2 + CE^2 = DE^2
CD^2 + (CE+EA)^2 = DA^2
CD^2 + CE^2 + 2CEEA + EA^2 = DA^2
DE^2 + 2CEEA + EA^2 = DA^2
CE=EA (diagonals bisect each other in parallelgram)
DE^2 + 2EA^2 + EA^2 = DA^2
therefore DE^2 + 3EA^2 = DA^2

To make it simpler for you at home to read:::

CD² + CE² = DE²
CD² + (CE+EA)² = DA²
CD² + CE² + 2CEEA + EA² = DA²
DE² + 2CEEA + EA² = DA²
CE = EA (diagonals bisect each other in parallelogram)
DE² + 2EA² + EA² = DA²
Therefore, DE² + 3EA² = DA²

Better?

 

[acmilan]

Hahaha good on you. you got it