Proof of a proposition relating rectangle and rhombus.

[blissful panther]

I tried to prove a proposition that states that by consecutively joining the midpoints of the sides of a rectangle, one obtains a rhombus. My strategy was to show that the 4 right triangles generated by this process are congruent by side, angle, side. Thus, it follows that segments EF, FG, GH, and HE are congruent. However, I realized that this wouldn't be enough to guarantee that this forms a rhombus, because firstly, there's nothing ensuring that this process creates a closed figure, nor that these segments (the supposed sides of a rhombus) are parallel to each other. I would have to ensure these two things as lemmas to be used in the proof. Am I being too pedantic? Can you see a more direct way to demonstrate the proposition in question without having to elaborate so much?

[sharp mesa]

<HEF=<HGF
<GHF=<GFE
If you prove this the problem will be solved

[blissful panther]

I've been thinking here for a while why I need to show both <HEF=<HGF and <GHE=<GFE. By showing only that <HEF=<HGF, I already ensure that the opposite sides are congruent. However, you are right, because to demonstrate that the opposite angles are also congruent, I also need to show that <GHE=<GFE. I also realized that I don't need to worry about ensuring that this process generates a closed figure, as I can simply say: "connect points E, F, G, and H to form the quadrilateral EFGH."

However, I'm not sure if this is enough to ensure that the opposite sides are parallel to each other, as I have a theorem here (let's call it T.15) already proven that guarantees if a quadrilateral is a parallelogram, then the opposite angles are congruent and the opposite sides are congruent. But here, I have that the opposite angles are congruent and the sides are also congruent, so I would have to appeal to the converse of T.15, which has not been demonstrated up to the exercise in question.

because, in my mind, I need to first show that this quadrilateral EFGH is a parallelogram, and furthermore, all its sides are equal (which we have already done) in order to then conclude that EFGH is a rhombus.

Why prove that <HEF=<HGF and <GHF=<GFE ensure that HG // EF and EH // FG?

[blissful panther]

So, as I have said, I have a theorem here that states that a parallelogram will have congruent opposite angles, but the converse of that has not been demonstrated. I would have to verify this first before affirming what you have said.

The converse that I would have to demonstrate would be: if a quadrilateral has congruent opposite angles, then this quadrilateral will be a parallelogram.

[blissful panther]

I was not sure about it, but it's true. I found this on web:

Since I don't have these theorems proven up to this point in the textbook I'm using, I won't be able to escape demonstrating one of them as a lemma. I was hoping there might be a shortcut, but it seems there isn't. I'll have to write several lines to finally conclude that EFGH is a rhombus.

[blissful panther]

So, you are right! It's a pity that I'll have to demonstrate a lemma first. 😅

Thanks @sharp mesa @iron siren ! 😄

[sharp mesa]

👍

[blissful panther]

@rain seal my problems was solved

I don't know how to change the status of this post from Unsolved to Solved 😅

[silent coral]

.solved