A grade 9 math question has me stumped

[Julie Smith]

This question has me stumped. 

I can easily figure out angle HEF is 70 degrees. But I don't know where to go after that. 

[Arcadia]

DS13 says EFGH is a parallelogram, due to points E, F, G, H being midpoints of their respective lines.

EH is 5.2 cm

angle EFG is 110 degrees 

ETA:

kid says to draw a line through AC

EF is parallel to AC is parallel to HG

or draw a line through BD to get

EH parallel to BD parallel to FG

Edited September 10, 2018 by Arcadia

[Julie Smith]

Thanks. 

Does anyone have any videos that will prove/ show how you can make the assumption that, "EFGH is a parallelogram, due to points E, F, G, H being midpoints of their respective lines"

 

[Kathy in Richmond]

I can't help with finding a video, but maybe I can explain why it works.

Draw segment AC in the figure. Look at triangle ABC, and notice that EF bisects two sides of this triangle (AB and BC)

By the Triangle Midsegment Theorem (most geometry courses include this, though it may have a different name in yours), EF must be parallel to the 3rd side, namely AC.

You can do the same thing with triangle ACD to show that HG is parallel to AC.

Since EF and HG are both parallel to AC, they must also be parallel to each other.

You can also draw segment BD and go through a similar argument to show EH and FG are both parallel to BD, and thus are parallel to each other.

Since both pairs of opposite sides are parallel, EFGH must be a parallelogram.

Edited September 10, 2018 by Kathy in Richmond
edited because I couldn't tell C from G :-P

[Arcadia]

36 minutes ago, Julie Smith said:

Does anyone have any videos that will prove/ show how you can make the assumption that, "EFGH is a parallelogram, due to points E, F, G, H being midpoints of their respective lines"

 

You might have missed the ETA that I added. Kid didn’t make an assumption. It was the conclusion to the ETA.

Kathy’s explanation is the same as my kid’s but hers is much clearer.