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.
enter image source here

1 Answer
Jim G.
Apr 29, 2018

#"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"#

Related questions
Impact of this question
1565 views around the world
You can reuse this answer
Creative Commons License