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In the given figure, $BH\perp AD$, $DC\perp AB$, $AF\perp BD$ and $DE=EB$ then prove that $AG$ is the diameter of the circle.

enter image description here

My Attempt:

$\angle AHB=90$ and $\angle ACD=90$. Let $BH$ and $CD$ intersects at $P$. Then, $\ ACPH$ is a cyclic quadrilateral.

Please, Help me to complete from here.

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    $PDGB$ is a parallelogram. You may use this to calculate $\angle ADG$. – 2017-01-06
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    Please elaborate, How is $PDGB$ a parallelogram? – 2017-01-06

2 Answers 2

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enter image description here

Let's call $P$ the orthocenter. Now we are going to prove that the symmetric point of $P$ w.r.t $E$ (let's call him $N$) lies on the circumcircle.

The quadrilateral $PBND$ is a parallelogram because both diagonals cut themself in the midpoint. Then $\angle DPB =\angle DNB$ but we know that $\angle DPB=\angle A -180º$ and so $\angle DNB=\angle A-180º$. That give us the quadrilateral $ABND$ is cyclic and this circle is, of course, the circumcircle.

Once $N$ and $G$ lies on the same line and on the same circle then $N=G$ and then $PBGD$ is a parallelogram. Now we have that $BH \parallel DG$ and once $BH \perp AD$ then $DG \perp AD$ what give us that $AG$ is the diameter of the circle.

P.S: Why $\angle DPB=\angle A -180º$?

In the triangle $AHB$ we have $\angle ABH = 90º-\angle A$ (becasue $\angle AHB =90º$). Then at the triangle $CPB$ we have $\angle CPB =\angle A$ (because $\angle PCB=90$ and $\angle CBP = \angle ABH = 90º-\angle A$) and $\angle CPB + \angle DPB = 180º$.

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    How is $E$ the mid point of $PG$? – 2017-01-06
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    It is a consequence of $N=G$. I built $N$ such that $E$ is the midpoint $PN$, after that I proved that $N$ lies on the circumcircle and as a consequence $N=G$ and then $E$ becames the mid point of $PG$. – 2017-01-06
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    @user354073: Is it clear now? – 2017-01-06
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    somehow. but not conpletely. Could you please explain in figure? – 2017-01-06
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    Is $G$ to be replaced by $N$? – 2017-01-06
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    @user354073: Yes, exactly! – 2017-01-06
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    In my circle, How do I show both the points $N$ and $G$? – 2017-01-06
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    You already have the point $G$ you don't need to show $N$ in your picture. Maybe should be good make another picture with just $N$ like I did and conclude that $N=G$ like I did. – 2017-01-06
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    Are $N$ and. $G$ coincident points – 2017-01-06
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    Yes, they are! I recommend you take a time understanding the solution! – 2017-01-06
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    Then, why is $\angle DPB=\angle A - 180$? – 2017-01-06
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    N and G lies on the same line. What does this mean? – 2017-01-06
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    E must be proved as the midpoint of PN **before** one can claim (via "diagonals bisecting each other") that PBND is a parallelogram. – 2017-01-06
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    @Mick: Nope! I built $N$ beeing the symmetric point of $P$ w.r.t. $E$. It means that $E$ is, by definition, the midpoint of $PN$. – 2017-01-06
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    OK. I think you mean the following:- Let N and G be two points on PE extended such that PE = EN and G is on the circle ABC as given. Since (1) $\angle DNB = \angle DPB$ (because they are opposite angles of //gm PDNB); and (2) $\angle DGB = 180^0 - \angle DAB = \angle HPC = \angle DPB$ (because of the cyclic quad AHPC and ADGB). From that, you conclude N is G. – 2017-01-07
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    @NeWtoN: are you good with that solution? – 2017-01-15
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    @Arnaldo, It is a bit confusing. Is there any other alternative? – 2017-01-16
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    @NeWtoN: maybe there is but I don't know. Perhaps you can try understand this solution because it is very constructive. – 2017-01-16
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    @Arnaldo, I did understand everything but I couldn't only understand two points – 2017-01-16
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    1. How is PDGB a parallelogram and – 2017-01-16
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    2. Why is E the mid point of PG? – 2017-01-16
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I have something close:

The point where the altitudes intersect is the orthocenter, lets call it $O$

I would like to define $G$ as the point of intersection of $OE$ and $AM$ and then show that that $G$ lies on the circle.

The point where the medians intersect is the centroid -- $N$ The center of the cirlce is the circumcenter -- $M$ enter image description here

$ME \parallel AF$ Not sure if I actually need this. But $M$ is at the intersection of the perpendicular bisectors of $ABC$

$O,N,M$ lie on a line called the Euler line

$ON:NM = 2:1$

$AN:NE = 2:1$

$N$ is also the centroid of $AGO$ $AO$ and $AE$ are medians of $AGO$ $ME$ is the midline of $AGO$

$AM = MG$

$G$ lies on the circle centered at $M.$

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    Could you please show me the figure for your proof? I could not get it from my figure. – 2017-01-06
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    @ Doug M, would you please give some time to explain this solution to me ? – 2017-02-27