Thread: Why are the two paralellograms equal in this construction?

1. Why are the two paralellograms equal in this construction?

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 (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.

2. 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.
Below your editor window you'll find "Additional Options" where you can attach images.

3. I apologize, first the topic title should say equal in area.

Q,R,J,K are defined as follows

Q = DE $\displaystyle \cap$ BI

R = FG $\displaystyle \cap$ CH

J = PA $\displaystyle \cap$ BC

K = PA $\displaystyle \cap$ IH

I will try to draw the picture in paint.

4. 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 (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? <--- h
what is the base of ABDE? <--- AB
what is the height of ABPQ? <--- h, because DE is the same straight line as PQ
what is the base of ABPQ? <--- AB
what is the height of JKIB? <--- d
what is the base of JKIB? <--- IB
...
Take PA as base of ABPQ. According to the text of the question $\displaystyle PA \parallel BI$ and PA and BI have the same length. Since IBQ are located on a parallel to PA the bas BI has the same height to JK as the base PA to BQ.
Thus the areas of ABPQ and BIJK must be equal.

I've attached a sketch and I hope now that I understood your description correctly.

5. Thank you so much for your help. I understand now. Sorry I couldn't get the picture to look nice at all in paint. Your posts were very helpful!