# Proof in Geometry. Is this solution correct?

Problem :

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!

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

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.