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The question is given as follows:

In the diagram below, $BF \perp HD$. Prove that $ACEG$ is a cyclic quadrilateral. Click here to see diagram

In class, we were told to introduce the origin $O$ and draw radii $OB, OH, OF,$ and $OD$. We then noticed that $\angle FOD = 180°- \angle FED$ and $\angle HOB = 180° -\angle HAB $, were told after that to use Thales' Theorem twice. At this point, I don't know where to use Thales' Theorem, or where to go with the problem.

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    Right, sorry. Edited. Thank you. – 2017-02-10

2 Answers 2

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Construction: Extend DO and HO to cut the circle at M and N respectively.

enter image description here

By considering the pink quadrilateral, we have the red marked angles are equal.

Similarly, the green marked angles are equal.

I forget the name of the following theorem (probably is called extended inscribed angle theorem). That fact is:-The size of the angle at X (HXB) is proportional to (arc HB + arc FD).

Since the angle at $X = 90^0$, then the proportional arc length is half the circumference = (arc MF + arc FD). This means arc HB = arc MF. Then, $\angle purple = \angle red$.

Result follows because $\angle green + \angle red =\angle green + \angle purple = 180^0$.

Added: Explaining "the theorem".

enter image description here

Let X be the point of intersection of chords AB and CD. It should be clear that:-

$\angle 3 = \angle 2 + \angle 1 = \angle2 + \angle 1’$

$\angle 2$ is proportional to the blue arc BD and $\angle 1’$ is proportional to the green arc AC. Hence, we get the required result.

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    I like it, but I don't quite get the "extended inscribed angle theorem". We haven't proved it in class. Is it an extension of Thale's Theorem? – 2017-02-10
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    @ColeMerkosky It is a rather simple theorem. See added. It has nothing to do with Thale's Theorem. Actually, the whole problem can be solved without using Thale's Theorem. – 2017-02-10
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    Gotcha. That makes everything clearer. Thanks! – 2017-02-10
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Here is another solution. Extend $AC$ and $GE$ until they intersect at a point $P$. Then since $AC$ and $GE$ are tangent to the circle at points $B$ and $F$ respectively, $PB = PF$ and therefore triangle $PBF$ is isosceles which means that $$\angle \, FBA = \angle \, FBP = \angle \, BFP = \angle\, BFG = \alpha$$ In the case when $AC$ and $GE$ do not intersect, they are parallel and then $$\angle \, FBA = \angle \, FBP = 90^{\circ} = \angle \, BFP = \angle\, BFG = \alpha$$ Analogously, we can conclude that $\angle \, AHD = \angle \, CDH = \beta$.

What follows is just angle chasing. Let $K$ be the intersection point of $HD$ and $BF$. In quad $ABKH$ the following angle identity holds $$360^{\circ} = \angle \, A + \angle \, ABK + \angle \, AHK + \angle \, BKH = \angle \, A + \alpha + \beta + 90^{\circ}$$ so $$\angle \, A = 270^{\circ} - \alpha - \beta$$ In quad $DEFK$ the following angle identity holds $$360^{\circ} = \angle \, E + \angle \, EFK + \angle \, EDK + \angle \, DKF = \angle \, A + (180^{\circ} - \alpha) + (180^{\circ} - \beta) + 90^{\circ}$$ so $$\angle \, E = \alpha + \beta - 90^{\circ}$$

Finally calculate $$\angle \, A + \angle \, E = 270^{\circ} - \alpha - \beta \, + \, \alpha + \beta - 90^{\circ} = 180^{\circ}$$ which holds exactly when the quadrilateral $ACEG$ is cyclic.