Geometry (circles and triangles) proofs

[Beam me down]

I'm having some trouble with one particular geometry proof:

From a point A outside a circle, a secant ABC is drawn cutting the circle at B and C, and a tangent AD touching at D. A chord DE is drawn equal in length to chord DB. Prove that triangles ABD and CDE are similar.

From that I've drawn the following:

http://img96.imageshack.us/img96/139/circle9we.gif

\angle ADB = \angle CED (as \angle ADB and \angle CED are alternant sements)
\angle CBD = 180 - \angle CED (1) (as they are opposite angles in a cyclic quadrilateral)

180 - \angle CBD = \angle DBA (2) as CBA is a straight line

\therefore \angle CED = \angle DBA substitute (1) into (2) This indicates that \angle BDA = \angle DBA. Is that right? And is that enough to prove that \bigtriangleup ABD \sim \bigtriangleup CDE ?

 

[VietDao29]

\angle ADB = \angle CED (as \angle ADB and \angle CED are alternant sements)

Nah, this is wrong.
\angle ADB \neq \angle CED
It should read:
\angle ADB = \angle ECD
(notice that the 2 chords DE = DB).
Can you see why? :)

 

[Beam me down]

Nah, this is wrong.
\angle ADB \neq \angle CED
It should read:
\angle ADB = \angle ECD
(notice that the 2 chords DE = DB).
Can you see why? :)

Is it because \angle ECB = \angle CDA = 90 which would make EC and DB parrelel in turn making \angle ADB and \angle ECD alternate angles?

I don't know how to prove this, but if \angle CDA = 90 that means CD is the diameter. No that has to be true as the radius (of which the diameter is an extension) meets any tangent at right angles, therefore the converse must be true.

If CD is the diameter could someone set me on the right track as to how to prove it?

 

[VietDao29]

Nah, it's wrong, since the problem does not state that EC and DB are parallel. You cannot assume that...
--------------
You should note that:
\angle ADB = \angle BCD (Tangent chord property)
\angle BCD = \angle ECD (DE = DB)
Can you go from here? :)

 

[GregA]

One more little hint if you need it...There is a segment on that circle that subtends an angle that you can find fairly easily...it also subtends a different angle

*edit* additionally, why is it assumed that the diameter of the circle connects the points D and C??...if you put C somewhere else you still get a secant ABC.

 

[Curious3141]

*edit* additionally, why is it assumed that the diameter of the circle connects the points D and C??...if you put C somewhere else you still get a secant ABC.

Who assumed that CD is a diameter ? It isn't in general.

 

[GregA]

I don't know how to prove this, but if \angle CDA = 90 that means CD is the diameter. No that has to be true as the radius (of which the diameter is an extension) meets any tangent at right angles, therefore the converse must be true.

If CD is the diameter could someone set me on the right track as to how to prove it?

That it isn't a diameter was also my thought...hence my reason for asking why it was assumed

 

[Curious3141]

That it isn't a diameter was also my thought...hence my reason for asking why it was assumed

Ah, I see. The orig. poster (Beam Me Down) was the only one who thought that (it's not correct).

I was actually unfamiliar with the tangent chord property that VietDao brought up (till he mentioned it and I googled for it).

The way I did this was in observing that

angle DEC = 180 - angle DBC (cyc quad) = angle ABD (supplementary angles).

Let's extend AD further and call a point on it F (A, D and F are collinear and D lies in between A and F). It should be obvious from the symmetry of DE = DB that angle EDF = angle BDA = theta (say). Then angle BDE = 180 - 2*theta. Angle BCE = 2*theta (opp angles, cyclic quad).

Now observe that angle DCE = angle DCB (equal chords ED and DB subtend equal angles). Therefore angle DCE = theta = angle BDA.

Since we've proven two corresponding angle pairs are equal (DEC = ABD & DCE = BDA), we're done.

In the process of doing this, one can also prove the tangent chord property.

 

[Beam me down]

I thought for a second that that I'd proved that \angle ADC was a right angle. But its definitely not in all cases.

 

[Curious3141]

I thought for a second that that I'd proved that \angle ADC was a right angle. But its definitely not in all cases.

You see how to do the problem now ?