Originally Posted by

**Sheld** Triangle ABC is an arbitrary triangle in $\displaystyle E^2$, not necessarily right. Draw parallelograms ABDE and ACFG on the outside of triangle ABC s.t. ABDE meets triangle ABC along the edge AB and ACFG meets triangle ABC along the edge AC. Let P be the point where the lines DE and FG meet. Draw a third parallelogram BCHI on the outside of ABC s.t. the vectors PA = BI.

I have the picture, but I am sorry I dont know how to Latex it to post it here.

My book says the the area of ABDE = area of ABPQ = area JKIB **<--- what is J, K, Q? and where?**(since they have the same base and height).

Looking at the picture, this makes no sense to me.

Can someone please tell me

what is the height of ABDE?

what is the base of ABDE?

what is the height of ABPQ?

what is the base of ABPQ?

what is the height of JKIB?

what is the base of JKIB?

and why they are equal?

I honestly don't understand why this is true by the picture. Thank you for your help.