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For reference:

In a convex quadrilateral ABCD (not convex in C) the extensions of sides BC and CD perpendicularly intersect sides AD and BC respectively. Calculate the measure of the angles which form the diagonals of the formed quadrilateral.

My progress:

$\triangle EDC \sim \triangle BFC (A.A.) \implies \measuredangle D = \measuredangle B\\ \measuredangle DCE =\measuredangle FCB = 90^\circ-\alpha\\ \measuredangle A = 2\theta\\ \measuredangle BCD = 2\alpha+2\theta\\ \measuredangle ECF = 180-(90-\alpha) = 90+\alpha =\measuredangle BCD \\ \triangle DFA: \alpha + 2\theta = 90^\circ\\ \triangle CEF (isosceles) :\measuredangle GEC = 45-\frac{\alpha}{2} = \measuredangle CEG$

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Something is missing to prove that the Angles are 90 degrees

Antonio
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peta arantes
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  • I don't think there's enough information... If you move B a little to the right, F changes location, so $\angle G$ changes as well. – Andria Aug 17 '21 at 22:36
  • Andria is correct - you may conclude that FEBD is cyclic and have $\alpha$ various values... – Moti Aug 18 '21 at 01:59

2 Answers2

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$\widehat {ECG}=\frac{(ECF=90+\alpha)}2=45+\alpha$

$\widehat {ECA}=\widehat {ECG}=45+\frac{\alpha}2$

$\widehat {EAC}=90-(45+\frac{\alpha}2)=45-\frac{\alpha}2$

Therefore:

$\widehat {ECG}=\widehat {AEG}$

that is triangles ECA and EGA are similar, since $\widehat {CEA}=90$, therefore :

$\widehat {EGA}=90^o$

sirous
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  • @sirous--can you explain $\widehat{ECA } = \frac{\widehat{ECG}}{2}?$ ..do you use the symbol "\widehat" for arc or angle? – peta arantes Aug 18 '21 at 17:25
  • Can I say that in the triangle ABD, C is orthocenter (meeting heights) so AC is perpendicular DB, angle CGE will measure 90 degrees because CG will be the height of triangle CEF? – peta arantes Aug 18 '21 at 23:56
  • @petaarantes, sorry for some errors, I corrected. You showed that triangle GEF is isosceles so quadrilateral BDEF is isosceles trapezoid so triangle ABD is alsp isosceles, therefore you can claim that C is orthocenter of triangle ABD. – sirous Aug 19 '21 at 04:44
  • @petaarantes, widehat is for angle. For arc is: \overset{\LARGE\frown}{FDI} which gives: $\overset{\LARGE\frown}{FDI}$. and means arc FI passing point D, for example. – sirous Aug 19 '21 at 06:34
  • @sirousgrateful for the clarifications..the widehat symbol i use for bow, hence the doubt...for angle i use \angle and \measuredangle – peta arantes Aug 19 '21 at 18:01
  • @ sirou$\angle EGC = \angle ECA = 45 +\frac{\alpha}{2}$ and not. $ 45+\alpha $ – peta arantes Aug 19 '21 at 19:50
  • @petaarantes, you are right, corrected. – sirous Aug 20 '21 at 03:53
  • grateful for the help – peta arantes Aug 24 '21 at 19:35
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Why did you stop? You are almost at the end. $$\angle DCF=180^{\circ}$$ $$\angle DCE+\angle ECF=180^{\circ}$$ $$\angle DCE+2\angle ECG=180^{\circ}$$(from triangle similarity) $$90^{\circ}-\alpha+2\angle ECG=180^{\circ}$$ $$\angle ECG=\frac{90^{\circ}+\alpha}{2}$$ Considering $\triangle ECG$, $$\angle ECG+\angle CEG+\angle EGC=180^{\circ}$$ $$\left(45^{\circ}+\frac{\alpha}{2}\right)+\left(45^{\circ}-\frac{\alpha}{2}\right)+\angle EGC=180^{\circ}$$ $$\angle ECG=90^{\circ}$$

ACB
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