Two circles intersecting, a lot of lines.

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The discussion revolves around the geometric properties of two intersecting circles, denoted as $T_1$ and $T_2$. The main objective is to demonstrate that if $\angle CDB = x$, then $\angle DEF = 2x$. Participants express confusion regarding the tangents drawn from points $A$ and $B$, and the order of points $A$, $B$, and $D$. The use of Geometer's Sketchpad (GSP) is recommended for visual experimentation, and it is concluded that the relationship holds true even when point $B$ is not positioned between points $A$ and $D$, provided the angles are interpreted correctly.

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caffeinemachine
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Let two circles $T_1$ and $T_2$, ($T_1$ is smaller than $T_2$), intersect at points $C$ and $E$. Let the tangent to $T_1$ at $C$ meet $T_2$ at $A$. From $A$ another tangent to $T_1$ is drawn which touches $T_1$ at $B$ and meets $T_2$ again at $D$. Let $F$ be the foot of perpendicular from $B$ to $CD$. $\angle CDB=x$. Show that $\angle DEF=2x$.

Drawing some figures to scale, I found that if $CD$ meets $T_1$ at $G$, then $\angle FEG=\angle GED$ although I am not able to prove this or solve the main question above.
 
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caffeinemachine said:
Let two circles $T_1$ and $T_2$, ($T_1$ is smaller than $T_2$), intersect at points $C$ and $E$. Let the tangent to $T_1$ at $C$ meet $T_2$ at $A$.

With you so far.

From $A$ another tangent to $T_1$

Did you mean a tangent to $T_{2}$? Or is $A=C$?

is drawn which touches $T_1$ at $B$

The tangent to $T_{2}$ at $A$ is by no means guaranteed to intersect $T_{1}$. Or is it postulated that it must?

and meets $T_2$ again at $D$. Let $F$ be the foot of perpendicular from $B$ to $CD$. $\angle CDB=x$. Show that $\angle DEF=2x$.

Drawing some figures to scale, I found that if $CD$ meets $T_1$ at $G$, then $\angle FEG=\angle GED$ although I am not able to prove this or solve the main question above.

I am confused, I'm afraid. Could you please help me out here?
 
Ackbach said:
With you so far.
Did you mean a tangent to $T_{2}$? Or is $A=C$?
The tangent to $T_{2}$ at $A$ is by no means guaranteed to intersect $T_{1}$. Or is it postulated that it must?
I am confused, I'm afraid. Could you please help me out here?
Okay let me upload a figure here.
 

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The picture does indeed help. However, I'm not sure I buy the result. Try drawing the figure again with hardly any overlap of the two circles. I end up with, for example, the points $ADB$ showing up in that order, not in the order of $ABD$. And the angle $\angle DEF$ sure looks a lot smaller than $\angle CDB$.
 
Ackbach said:
The picture does indeed help. However, I'm not sure I buy the result. Try drawing the figure again with hardly any overlap of the two circles. I end up with, for example, the points $ADB$ showing up in that order, not in the order of $ABD$. And the angle $\angle DEF$ sure looks a lot smaller than $\angle CDB$.
I guess you are right. I don't know what's going on here. Perhaps an additional constraint should be imposed... that ABD occur in this order only.
 
Try constructing the drawing in something like Geometer's Sketchpad, and see if you can find anything out. GSP allows lots of experimentation by dragging things around and generally having fun.
 
Ackbach said:
Try constructing the drawing in something like Geometer's Sketchpad, and see if you can find anything out. GSP allows lots of experimentation by dragging things around and generally having fun.
Thank you for telling me about this software. I will download its trial version as soon as I can. Meanwhile I checked(using ruler and compass :() that if B is between A and D then the problem makes sense.
 
caffeinemachine said:
Thank you for telling me about this software. I will download its trial version as soon as I can. Meanwhile I checked(using ruler and compass :() that if B is between A and D then the problem makes sense.
I have not made any progress with this problem, but I suspect that the result is still true when B is not between A and D, provided that you interpret it correctly. In that case, you should take the angle $x$ to be $\angle CDA$, which will be obtuse. Then $2x$ will be greater than $180^\circ$ and you need to take the reflex angle $\angle DEF$.
 

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