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$△ABC$ is an isosceles right-angled triangle with $∠A = 90^\circ$. $DE\parallel BC$. Square $DFGH$ is constructed, with $F$ lying on $AC$ and $G$ on $BC$. Prove that $∠EDF=∠EGF$.

I attempted to prove it using some equivalent methods, such as demonstrating the concyclicity of points DEFG or the perpendicularity of EG to BC, etc. However, I couldn't find a method to prove it, or I kept falling into the fallacy of circular reasoning.

Appreciate any hints or assistance in solving the problem.

uggupuggu
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Alibuda
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    Please edit the post to include what you have tried so far and where you got stuck. Or else your question will likely be downvoted and closed. – Haris Feb 15 '24 at 07:04
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    New to here, thanks for the reminder. I can only attempt some problem-solving approaches reluctantly, as I am truly clueless. – Alibuda Feb 15 '24 at 07:19
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    I'm curious how you tried to show that $DEFG$ is cyclic. Showing is better than telling. – David K Feb 15 '24 at 07:20
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    You might find the math a little easier to write if you use MathJax. I see someone has already shown how to do it by editing your question. – David K Feb 15 '24 at 07:21
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    Hint: $\angle DGF = 45^\circ= \angle DEA$ – D S Feb 15 '24 at 07:27
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    Concyclicity of DEFG is a good approach. – Aig Feb 15 '24 at 07:36
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    Thank you guys! ∠DGF=45∘=∠DEA, so ∠DGF + ∠DEC = ∠DEA + ∠DEC = 180∘. Proves the concyclicity of DEFG. – Alibuda Feb 15 '24 at 07:48
  • To complete the proof: ∠EFG + ∠EDG = (∠EFD + 90∘) + (∠EFD + 45∘) = (∠EFD+ ∠EFD) + 135∘ = ∠AED + 135∘ = 45∘+135∘ = 180∘ – Alibuda Feb 15 '24 at 08:06
  • Please, put your point $F$ further from $E$ in order to improve visibility. – Dominique Feb 15 '24 at 08:13
  • A square is rectangle and a rectangle is a parrallelgram with right angles. Use the parallell line, interior angle postulate and just just chase the thing into the ground. It's not circular. – fleablood Feb 15 '24 at 16:58
  • "or the perpendicularity of EG to BC," No-one said $EG$ was perpendicular to $BC$. – fleablood Feb 15 '24 at 17:07
  • "Square DFGH is constructed" this implies a square can be contstructed. Maybe go through to motions of constructing a square. The me it's obvious that $\angle EDF\cong \angle HGB$ (because $DE || HG$ and $DG || FG$) and that $\angle HGB \cong \angle EGF \iff \angle FGH \cong \angle EGH$ (which is right). So it is suffecient to prove $EG \perp BC$ and maybe that was part of the construction. – fleablood Feb 15 '24 at 18:35
  • Is this an 8th grade problem? Maybe an 8th grade olympiad geometry problem? – Starlight Feb 16 '24 at 03:12
  • Alibaba, why did you mix up the letters in your answer. – Bob Dobbs Feb 20 '24 at 20:18

2 Answers2

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Here is the solution based on the helpful tips from everyone, and I consider it as my practice using MathJax. For greater clarity, I have re-labeled the names of the points in the image. enter image description here

Problem:
$△ABC$ is an isosceles right-angled triangle with $∠A = 90^\circ$. $GH\parallel BC$. The three points of square $GDEF$ lie on the three sides of $△ABC$. Prove that $∠8=∠5$.

Solution:
The objective is to demonstrate that the quadrilateral $GHDE$ is concyclic, thereby proving that $∠8=∠5$.
If it can be proven that the sum of both pairs of opposite angles in the quadrilateral $GHDE$ is $180^\circ$ each, then the four points lie on a circle. Furthermore, since $∠4 = 90^\circ$, it can be deduced that $GE$ is the diameter of this circle, and $∠1 = 90^\circ$; it follows that $HE$ is perpendicular to both $GH$ and $BC$.
Sum of opposite angles in the blue group: $(∠1+∠2) + (∠5+∠6) = 135^\circ + 45^\circ = 180^\circ$.
Sum of opposite angles in the orange group: $(∠3+∠4) + (∠7+∠8) = (∠3+90^\circ) + (45^\circ+∠8)$ $= (∠3+∠8) + 135^\circ = ∠9 + 135^\circ = 45^\circ + 135^\circ = 180^\circ$.
Q.E.D.

Alibuda
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enter image description here

Draw a line passing H and parallel with DE. Extend ED to get point Q such that EQ is equal to distance between E and the line.Construct a square on side EQ. Draw a circle center at O, the center of square and radius OE, it intersects AC at F and AC at G. Construct square DFGH on side DE. In this way we get :

$\angle EDF=\angle EGF$

because they are both opposite the arc EF.

sirous
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    Thanks for your work. How do you know that DF will intersect FG vertically? (In "...it intersects AC at F and AC at G," is the second occurrence of AC intended to be BC?) – Alibuda Feb 16 '24 at 09:30
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    @Alibuda, By this construction I wanted to show that the square DFEH is in fact the result of rotation of square QEIH, that is why DF is perpendicular at FG at F. Certainly this is an optimized construction, ie, BC has a certain location which relates to the dimension of the square QEIH., in other words the distance between DE and BC defines the measure of the side of the square. So we may have various constructions, but they are all the result of rotation of square QEIH is different sizes. – sirous Feb 16 '24 at 15:03