Folding Paper and Angles Question

Question

We are trying to solve this problem asking us to look for a certain angle involving a paper being folded.

Here's the problem:

On a square-shaped paper ABCD, folds the paper according to line EF to make an equal half. Again, generates ab equal half ti each side of the paper, where A and B meet each other. As shown in the image below, what is X?

X° in our solution results in 45° but in the answer key it's 75° and we can't seem to find the reason why. Thanks in advance!

Answer 1

Let's analyse your working:

"Hi, thanks for your reply. We began our solution by assuming AD = CD, and this makes angle GDC 60°. Let's say that midpoint between A and E is H. Isn't angle EHG also 60° ? Then, we also got angle ADH as 15° so angle AHD is 75°. X° is then 180° - 60° - 75° = 45°."

1. Because $AC$ = $CD$, after folding the corners, you obtain an equilateral triangle, so $\angle GDC$ = $60^\circ$ as you said.
2. We denote the midpoint of $AE$ by $H$. What is $\angle EHG$? First, from 1., we see that $\angle GDF = 30^\circ$. We also know that $\angle GDH = 90^\circ$, as the original paper is square. Using "angles on a line add to $180^\circ$", we have $\angle EGH = 180^\circ-90^\circ-30^\circ = 60^\circ$. The sum of angles in a triangle add to $180^\circ$, so $\angle EHG = 180^\circ - 60^\circ - 90^\circ = 30^\circ \neq 60^\circ$.
3. The rest of your argument is fine.

In summary, you need to calculate $\angle EHG$ more carefully. You'll see that everything works out now.

Answer 2

Here you can see, The answer is $75^°$.