Welcome to our community

Be a part of something great, join today!

Two circles intersecting, a lot of lines.

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
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.
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
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?
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
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.
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
IMG_0170.PNG
I think this should make things clear.
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
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 $\angle CDB$.
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
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.
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
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.
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
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.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
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 $\angle DEF$.