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- #1

- Mar 10, 2012

- 835

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.

- Thread starter caffeinemachine
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- Thread starter
- #1

- Mar 10, 2012

- 835

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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- #2

- Jan 26, 2012

- 4,198

With you so far.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 $A$.

Did you mean a tangent to $T_{2}$? Or is $A=C$?From $A$ another tangent to $T_1$

The tangent to $T_{2}$ at $A$ is by no means guaranteed to intersect $T_{1}$. Or is it postulated that it must?is drawn which touches $T_1$ at $B$

I am confused, I'm afraid. Could you please help me out here?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.

- Thread starter
- #3

- Mar 10, 2012

- 835

Okay let me upload a figure here.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?

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- #4

- Mar 10, 2012

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- #5

- Jan 26, 2012

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- Thread starter
- #6

- Mar 10, 2012

- 835

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.

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- #7

- Jan 26, 2012

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- #8

- Mar 10, 2012

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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.

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- #9

- Feb 7, 2012

- 2,770

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 $\angle DEF$.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.