Problem :

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X, Z, P, F, and G were chosen by me to form other triangles which go through the center of the circle. I have picked the center by guessing. Please let me know if there is a concrete way of finding the center of a circle. $$ Show: AE \times BE=CE \times DE \\ \text{proof of similarity: } \\ \angle CDA = \angle CDG - \angle ADP \\ = \frac{1}{2} \angle FGD - \frac{1}{2} \angle AGP \\ = \frac{1}{2} \angle CGA \\ \angle ABC = \angle ABG - \angle CGB \\ = \frac{1}{2} \angle ZGB - \frac{1}{2} \angle BGF \\ = \frac{1}{2} \angle ZGF = \frac{1}{2}CGA \\ \text{this proof that } \angle CDA = \angle ABC \\ \text{On to proving the next angle similarity:} \\ \angle DCB = \angle DCG + \angle GCB \\ = \frac{1}{2}\angle CGP + \frac{1}{2} \angle FGB \\ =\frac{1}{2} \angle DGB \\ \angle DAB = \angle BAG + \angle GAD \\ = \angle \frac{1}{2} XGA + \frac{1}{2} \angle PGA \\ = \frac{1}{2} \angle DGB \\ \text{So } \triangle CDG \text{ is similar to } \triangle AGB \\ \text{Finally: for the triangle outside the circle.} \\ \angle EDB = \angle EBD \\ \text{there for } \triangle DEB \text{ is isosceles.} \\ \text{ Note that I assume here that I do not have to prove: } \angle ADB = \angle CBD \rightarrow \text{ I assume that it is valid to say: } \angle EDB = \angle EBD \text{ because I have proven } \\ \angle CDA = \angle ABC \text{ this might be wrong though. } \\ \text{Now I know:} \triangle CEA \text{ is similar to } \triangle DEB \\ \\ \text{now I show that: } \\ AE \times BE=CE \times DE \\ \frac{CE}{EA} = \frac{BE}{DE} \rightarrow EA \times BE = CE \times DE \\ \text{I conclude: This proves the problem.} $$

I only proved two angles because I remember this being enough to proof similarity. If I am mistaken, please do say.

I you know of any books that have problems like this, please let me know. I really enjoy working with these.

Thank you!

  • $\begingroup$ To find the center, construct two chords, construct perpendiculars to their midpoints. The perpendiculars will intersect at the center. $\endgroup$ – John Wayland Bales Jan 1 '17 at 18:24
  • $\begingroup$ @JohnWaylandBales Thank you, that explains that part. $\endgroup$ – Cro-Magnon Jan 1 '17 at 18:26

1) What you want to show and the conclusion you get did not reflect your aim of “finding the center”.

2) Showing the said products are equal can be done by proving $\triangle CEB \sim \triangle AED$ via the “angle in the same segment theorem”. That means $\angle B = \angle D$ and $\angle ECB = … = \angle EAD$. Result follows directly from the equality of the corresponding ratios.

3) What you are trying to prove is commonly known as “the power of a point”.

  • $\begingroup$ You're right. I did not state the objective clearly enough. Your second point is what I wanted to ask. Finding the center was just a side question. I will adjust the question later. Thank you. $\endgroup$ – Cro-Magnon Jan 2 '17 at 10:12
  • $\begingroup$ Also, do you know how I can practice this better? I always seem to take the wrong or unintuitive/smart path to finding the answer.. Any books you would reccommend? $\endgroup$ – Cro-Magnon Jan 2 '17 at 10:15
  • $\begingroup$ @Cro-Magnon Search the internet using "the power of a point" as key. I am sure you will find lots of useful reference material. $\endgroup$ – Mick Jan 2 '17 at 10:18

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