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Let $ABCD$ be a right-angled trapezium with $BC \parallel AD$ and $CD \perp AD, BC$. Let $\Gamma_A$ be the circle centered at $A$ with radius $AD$, and let $\Gamma_B$ be the circle centered at $B$ with radius $BC$. Suppose $\Gamma_A$ and $\Gamma_B$ intersect at points $E$ and $F$. Let $\omega$ be a circle tangent internally to $\Gamma_A$ at $G$ and to $\Gamma_B$ at $H$. Prove that the lines $GD$, $EF$, and $HC$ are concurrent.

We could use radical axis such that for any three circles, their pairwise radical axes are concurrent at a single point (the radical centre). To use this here, we could identify three specific circles whose radical axes correspond to the lines $GD$, $EF$, and $HC$.

The line $EF$ is clearly the radical axis of $\Gamma_A$ and $\Gamma_B$. To show that $GD$ and $HC$ are also radical axes, we might consider the existence of a phantom circle $\Sigma$ that passes through the four points $\{C, D, G, H\}$. If such a circle exists, $GD$ would be the radical axis of $\Gamma_A$ and $\Sigma$, while $HC$ would be the radical axis of $\Gamma_B$ and $\Sigma$.

Otherwise, we could consider homothety, which is another idea.

Is there a solution following either?

Regards

enter image description here

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    $\begingroup$ Your diagram does not match the description of the problem. $\endgroup$ Commented Feb 13 at 9:21
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    $\begingroup$ I think you may draw a new picture including the points $G$ and $H$ (by drawing the three circles first). $\endgroup$ Commented Feb 13 at 9:21
  • $\begingroup$ $BC$ doesn't look parallel to $AD$ in the figure. $\endgroup$ Commented Feb 13 at 9:35
  • $\begingroup$ You may copy my diagram into your question. $\endgroup$ Commented Feb 13 at 9:47
  • $\begingroup$ Can we attempt your problem via coordinate geometry? $\endgroup$ Commented Feb 13 at 11:22

3 Answers 3

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Not an answer, just providing a correct diagram. Your idea that $CDHG$ is cyclic is correct. We also should note that the concurrence is also a point on the tangent circle $\omega$, whose center I have denoted $O$.

The fact that $ABCD$ is a right trapezium implies that $CD$ is the common external tangent line to the circles $\Gamma_A$, $\Gamma_B$. It is also clear that $EF \perp AB$.

enter image description here

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  • $\begingroup$ is that geogebra? how does one make circle tangent to 2 circles in it? $\endgroup$ Commented Feb 14 at 9:24
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    $\begingroup$ @NooneAtAll3 Yes, the figure was made in Geogebra. To draw $\odot O$ (red circle), one needs to utilize the fact that the locus of centers of such tangent circles is a hyperbola whose foci are $A$ and $B$. In Geogebra, you would use the hyperbola tool to create this locus, and after selecting the foci, select either one of the intersection points $E$, $F$ as a point on the hyperbola. Then $O$ is allowed to be any point on this hyperbola. $G$ is collinear with $OA$ and $H$ is collinear with $OB$. $\endgroup$ Commented Feb 14 at 9:31
  • $\begingroup$ {+1] Good idea to use this hyperbola. I mention this locus in the extension I propose here. $\endgroup$ Commented Feb 14 at 21:35
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Here's a phantom point approach.

Consider a circle passing through $C$,$D$, and $H$. Let it intersect $\Gamma_A$ at $G'$. By the radical axis theorem, $CH$, $EF$ and $DG'$ concur. Let $L$ be the point of intersection. If we show that $G = G'$, we are done.

This is easy. Consider: $$\measuredangle LHG' = \measuredangle CHG' = \measuredangle CDG' = \measuredangle DEG$$ This shows that the tangent at $G'$ to $(LHG')$ and $\Gamma_A$ coincide. Similarly, it is shown that the tangent at $H$ to $(LHG')$ and $\Gamma_B$ coincide. Thus, $(LHG')$ is tangent to both $\Gamma_A$ and $\Gamma_B$. So infact, $G$ must be the same as $G'$, and we are done.

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enter image description here

This picture shows clearly the way of construction and finding the tangent points G and H. To draw circle $\omega$ we need the center O which is the intersection of segments connecting the centers of $\Gamma_A$ and $\Gamma _B$ to the tangent points G and H respectively.Another way is to use a third point which can be the intersection of DG and CH . Tangent from C and W to circle $\omega$ meet at point U , with respect to $\omega$ we have:

$UG= UH$

This means that U is on the radical axis of circles $\Gamma_A$ and $\Gamma_B$. Also we have:

$WH= CG\Rightarrow UW= UC$

Clearly UZ is the altitude of isosceles triangle UCW and the center of circle $\omega$ is on it.In triangle CDW we have $ZQ||WD$ which means Q is the midpoint of DC. This is a property of the radical axis of two intersecting circles, that is Q is also is on the radical axis . In the way UQ is coincident on radical axis of $\Gamma_A$ and $\Gamma_B$.Since $HW= CG$ and WH is the transformation of DG we may write:

$CI\times CH =DI\times DG$

This means I is symmetric for circles $\Gamma_A$ and $\Gamma_B$ so it is on their radical axis.

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