MHB Area congruence in two quadrilaterals

[My Name is Earl]

Using the theorems:
Area of a triangle = (1/2)bh
Area of a parallelogram = bh
Area of a trapezoid = (1/2)h(b₁+b₂)
Solve the following:

View attachment 8517

 

[skeeter]

Hint ...

triangle AED contains half the area of both given parallelograms

 

[My Name is Earl]

Hint ...

triangle AED contains half the area of both given parallelograms

That is true if you assume quadrilateral EFGA is a parallelogram. Only quadrilateral ABCD is defined as a parallelogram. You would have to prove that EFGA is a parallelogram, then the problem becomes easy using area formulas. How do you prove that EFGA is a parallelogram with the given information?

 

[I like Serena]

That is true if you assume quadrilateral EFGA is a parallelogram. Only quadrilateral ABCD is defined as a parallelogram. You would have to prove that EFGA is a parallelogram, then the problem becomes easy using area formulas. How do you prove that EFGA is a parallelogram with the given information?

We can't prove that EFGA is a parallelogram.
It could be any type of quadrilateral, and we can't say anything about its area either.
Note that we can pick F anywhere we want, and after picking F we can pick G anywhere on the line DF.

In other words, it seems that the problem statement should have said that EFGA is a parallelogram, and that it was accidentally left out.

 

[Greg]

I find it strange that there is no reference to the position of point $D$.

 

[My Name is Earl]

We can't prove that EFGA is a parallelogram.
It could be any type of quadrilateral, and we can't say anything about its area either.
Note that we can pick F anywhere we want, and after picking F we can pick G anywhere on the line DF.

In other words, it seems that the problem statement should have said that EFGA is a parallelogram, and that it was accidentally left out.

I spent days on this one to no avail. I can pick an E such that AE is parallel to GF, but that does not allow me to show that AE = GF. I really believe the author meant EFGA not to be initially defined as a parallelogram. I sure do wish I could see their solution to this one. Thanks for the reply!

 

[My Name is Earl]

I spent days on this one to no avail. I can pick an E such that AE is parallel to GF, but that does not allow me to show that AE = GF. I really believe the author meant EFGA not to be initially defined as a parallelogram. I sure do wish I could see their solution to this one. Thanks for the reply!

I contacted the author of the book. To my surprise I got a quick response. He stated that the problem is not solvable without saying that EFGA is a parallelogram. There was an error during the printing process and the problem was supposed to state this fact. Problem solved!