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What we know is that: 1. ABCD is a quadrilateral. 2. The red area is a square. 3. AH=BE=CF=DG

enter image description here

The question is prove that ABCD is also a square. I have realised that the four triangles here AHG, DGF, EFC and HBE have the same length hypotenuse and also AH = DG = CF = BE, so if I can prove ∠ A, B, C, D are 90°, then four triangles are congruent. Then I will know that four sides, AB,BC,CD,DA are the same length then I can prove it. The problem is that I dont know how to prove angle A,B,C,D are 90 degree. Thanks!

NickMan
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4 Answers4

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If one among the angles at $A$, $B$, $C$, $D$ is a right angle, then it is easy to prove they are all right angles and $ABCD$ is a square. Suppose then none of them is a right angle: at least one of them ($\angle GAH$, for instance) must then be obtuse. I'll show that this leads to a contradiction.

Let $N$ be the foot of the perpendicular from $G$ to line $AH$. As $\angle GAH>90°$ then $AH<NH$.

Let $M$ be the foot of the perpendicular from $E$ to line $BH$: we have then $EM\le EB$. But triangles $EMH$ and $HNG$ are congruent (because they have $\angle NGH=\angle MHE=\pi/2-\alpha$, $\angle NHG=\angle MEH=\alpha$ and $GH=HE$), thus: $$ AH < NH = EM \le EB $$ which is in contradiction with the hypothesis $AH=EB$.

enter image description here

Intelligenti pauca
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  • Thanks for your thoughts! You said triangles EMH and HNG are congruent because they have the same angles 90° and EH=GH... If the hypotenuse and one leg of a right triangle are equal to the hypotenuse and one leg of another right triangle, then the two right triangles are congruent. EH=GH but what is the one leg of the triangle ? – NickMan Sep 03 '17 at 21:40
  • "They have the same angles" means all their angles are equal pairwise: $\angle NGH=\angle MHE = 90°-\alpha$, and so on. – Intelligenti pauca Sep 04 '17 at 15:09
  • In other words: triangles $NGH$ and $BHE$ have $\angle NGH=\angle MHE$, $\angle NHG=\angle MEH$ and $GH=HE$. They are then congruent by ASA. – Intelligenti pauca Sep 07 '17 at 07:22
  • Thank you for your explaination, now I understand...you provide the very first solution that I can totally understand under this question. People were trying to prove triagles FCE=EBH=HAG=GDF but i just dont see how this could be done. Thanks for your thoughts! – NickMan Sep 08 '17 at 14:04
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HINT: prove that the triangles $$FCE=EBH=HAG=GDF$$ are congruent

Dr. Sonnhard Graubner
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Hint: extend side $AD$ and $EF$. Let's say they intersect at point $X$. Then you can see that $\angle{XFG}=90°$ and $\angle{DXF}\cong\angle{AGH}$. Can you take it from here?

Vasili
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  • Thanks for the hint but I still have no idea how to proceed ... If I can prove triangle AHC and triangle CXF are similar, then I can say ∠A = ∠XFG=90°... but how? – NickMan Sep 02 '17 at 17:51
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Scheme of the proof:

  • If $ABCD$ is rectangle, then it is easy to show that it is a square too (congruent triangles).
  • If $ABCD$ isn't rectangle, then one of angles $A,B,C,D$ is obtuse one. W.l.o.g., $A>90^{\circ}$.
    • Then the distance from the line $AB$ to point $E$ is greater than $|AH|$.
    • So, $|BE|>|AH|$; contradiction.

More detailed:

first, draw the case with right angle $\angle A$:

Illustration 1

note that:

  • point $A$ is on the semicircle $GAH$ (with diameter $GH$);
  • line $GA$ is tangent to orange circle;
  • line $AB$ is tangent to blue circle;
  • $|AH|=|BE|$.

Now, build point $A'$ such that $|AH|=|A'H|$ and $\angle GA'H>90^{\circ}$:

Illustration 2


Point $A'$ belongs to orange circumference, and is inside the semicircle $GAH$ (inside $\triangle GAH$ too).
So $\angle A'HG<\angle AHG$, hence ray $A'H$ cannot intersect blue circle.
So any segment $B'E$ (where $B'$ belongs to the ray $A'H$) has length greater than $|BE|$.

Here we came to contradiction: if $\angle GA'H$ is obtuse, then $|B'E|>|A'H|$.

Oleg567
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  • Thanks for your thoughts. But how I can prove these four triangles are congruent? If I can prove the ∠A = ∠B= ∠C=∠D = 90° then it would be easy. – NickMan Sep 03 '17 at 08:49
  • If assume that $ABCD$ isn't rectangle, then we have contradiction: $|EB|>|AH|$. Therefore, $ABCD$ must be rectangle (otherwise we will have contradiction), and (see p.1) hence is a square. – Oleg567 Sep 03 '17 at 09:46
  • If we use the picture from Aretino then the distance from the line AB to point E is ME, how do you claim ME is greater than AF? – NickMan Sep 04 '17 at 09:37
  • @NickMan, I updated my answer with illustrations now. Hope it is helpful. – Oleg567 Sep 05 '17 at 06:45
  • @NickMan, my fault... I had to say " then $ME$ is greater than $AH$" ($AH$, not $AF$). Now all is corrected. – Oleg567 Sep 05 '17 at 07:39
  • Thanks for making the drawing and explanation! So first, i draw semicircle GAH (with diameter GH), then I will find a point A on the semicircle so that it becomes a right angle ∠A. Then if i draw a orange circle with AH as radius, then line GA is tangent to orange circle. Then I extend this AH to AB, where ∠ABE is 90°.Then the line AB is tangent to blue circle whose radius is BE. Now my questions is : how do you know |AH|=|BE|? – NickMan Sep 08 '17 at 13:55
  • @NickMan, I place point B at the distance $|AG|$ from the point $H$. Then $\triangle AHG = \triangle BEH$ (since $|GA|=|HB|$, $|GH|=|HE|$ and $\angle BHE= 180^{\circ}-\angle GHE -\angle AHG = 180^{\circ}-90^{\circ}-\angle AHG = \angle AGH$). – Oleg567 Sep 13 '17 at 15:13