I am solving some past exam questions for a japanese university entrance exam and got stuck in this question:
In this pyramid, each side of the square $ABCD$ equals $2a$, also the height $OH = a$, $M$ is a mid point on $AB$.
Finally $AE$ and $OB$ are orthogonal, $CE$ and $OB$ are also orthogonal.
I found out $HM$ = $a$
$OM$ = $\sqrt2 a$
$OB$ = $\sqrt3 a$
Now we have to find the area o $AEC$, but first we need $AE$ and this is where i got stuck, i know that angle $AEB$ = $90°$, since we know $AB$ = $2a$, we just need $EB$ to find $AE$, but i can not see how to find it, is probably very simple, so sorry in advance, thank you.
Since OH is the height of the pyramid, the pyramid is a right one (i.e. with all the side edges are equal).
OH is known, and HC can be found imply OB = OC can be determined.
Get $AE$ from $OM \times AB = OB \times AE$.