Introduction, circle geometry

Let $\widehat{OEC}=x$ .

• $\widehat{COE}=\widehat{CBF}$
• $\widehat{ECO}=\widehat{FCB}$
• $\widehat{OEC}=\widehat{CFB}$

1. This relation reminds us of a proportionality relation between similar triangles. So investigate which triangles contain these sides and prove them similar. In this case 3 equal angles works well. Start with one triangle.

   In $▵EDC$

   $$\begin{array}{ccc}\hfill \widehat{CED}& =& {180}^{\circ }-x\left(\angle \mathrm{\text{'}s\; on\; a\; str.\; line}\mathrm{AD}\right)\hfill \\ \hfill \widehat{ECD}& =& {90}^{\circ }-x(\mathrm{complementary}\angle \mathrm{\text{'}s})\hfill \end{array}$$
2. Now look at the angles in the other triangle.

   In $▵ADC$

   $$\begin{array}{ccc}\hfill \widehat{ACE}& =& {180}^{\circ }-x(\mathrm{sum\; of}\angle \mathrm{\text{'}s}\widehat{\mathrm{ACE}}\mathrm{and}\widehat{\mathrm{ECO}})\hfill \\ \hfill \widehat{CAD}& =& {90}^{\circ }-x(\mathrm{sum\; of}\angle \mathrm{\text{'}s\; in}▵\mathrm{CAE})\hfill \end{array}$$
3. The third equal angle is an angle both triangles have in common.

   Lastly, $\widehat{ADC}=\widehat{EDC}$ since they are the same $\angle$ .

4. Step 4d: Now we know that the triangles are similar and can use the proportionality relation accordingly.
   $$\begin{array}{ccc}& \therefore & ▵ADC|||▵CDE(3\angle \mathrm{\text{'}s\; equal})\hfill \\ & \therefore & \frac{ED}{CD}=\frac{CD}{AD}\hfill \\ & \therefore & C{D}^2=ED.AD\hfill \end{array}$$

1. This looks like another proportionality relation with a little twist, since not all sides are contained in 2 triangles. There is a quick observation we can make about the odd side out, $OE$ .
   $$\begin{array}{ccc}\hfill OE& =& CD(▵\mathrm{OEC}\mathrm{is\; isosceles})\hfill \end{array}$$
2. With this observation we can limit ourselves to proving triangles $BOC$ and $ODC$ similar. Start in one of the triangles.

   In $▵BCO$

   $$\begin{array}{ccc}\hfill \widehat{OCB}& =& {90}^{\circ }(\mathrm{radius}\mathrm{OC}\mathrm{on\; tangent}\mathrm{BD})\hfill \\ \hfill \widehat{CBO}& =& {180}^{\circ }-2x(\mathrm{sum\; of}\angle \mathrm{\text{'}s\; in}▵\mathrm{BFC})\hfill \end{array}$$
3. Then we move on to the other one.

   In $▵OCD$

   $$\begin{array}{ccc}\hfill \widehat{OCD}& =& {90}^{\circ }(\mathrm{radius}\mathrm{OC}\mathrm{on\; tangent}\mathrm{BD})\hfill \\ \hfill \widehat{COD}& =& {180}^{\circ }-2x(\mathrm{sum\; of}\angle \mathrm{\text{'}s\; in}▵\mathrm{OCE})\hfill \end{array}$$
4. Again we have a common element.

   Lastly, $OC$ is a common side to both $▵$ 's.

5. Step 5e: Then, once we've shown similarity, we use the proportionality relation , as well as our first observation, appropriately.
   $$\begin{array}{ccc}& \therefore & ▵BOC|||▵ODC(\mathrm{common\; side\; and\; 2\; equal\; angles})\hfill \\ & \therefore & \frac{CO}{BC}=\frac{CD}{CO}\hfill \\ & \therefore & \frac{OE}{BC}=\frac{CD}{CO}(\mathrm{OE}=\mathrm{CD}\mathrm{isosceles}▵\mathrm{OEC})\hfill \end{array}$$

Let $\angle BCD=x$

1. To show these 2 triangles similar we will need 3 equal angles. We can use the result from the previous question.

   Let $\angle FEA=y$

   $$\begin{array}{ccc}& \therefore & \angle FDA=y\left(\angle \mathrm{\text{'}s\; subtended\; by\; same\; chord}\mathrm{AF}\mathrm{in\; cyclic\; quad}\mathrm{FADE}\right)\hfill \\ & \therefore & \angle CBD=y(\mathrm{corresponding}\angle \mathrm{\text{'}s,}\mathrm{FD}\parallel \mathrm{CB})\hfill \\ & \therefore & \angle FEA=\angle CBD\hfill \end{array}$$
2. We have already proved 1 pair of angles equal in the previous question.
   $$\begin{array}{ccc}\hfill \angle BCD& =& \angle FAE\left(\mathrm{above}\right)\hfill \end{array}$$
3. Proving the last set of angles equal is simply a matter of adding up the angles in the triangles. Then we have proved similarity.
   $$\begin{array}{ccc}\hfill \angle AFE& =& {180}^{\circ }-x-y\left(\angle \mathrm{\text{'}s\; in}▵\mathrm{AFE}\right)\hfill \\ \hfill \angle CBD& =& {180}^{\circ }-x-y\left(\angle \mathrm{\text{'}s\; in}▵\mathrm{CBD}\right)\hfill \\ & \therefore & ▵AFE|||▵CBD(3\angle \mathrm{\text{'}s\; equal})\hfill \end{array}$$

1. This equation looks like it has to do with proportionality relation of similar triangles. We already showed triangles $AFE$ and $CBD$ similar in the previous question. So lets start there.
   $$\begin{array}{ccc}\hfill \frac{DC}{BD}& =& \frac{FA}{FE}\hfill \\ & \therefore & \frac{DC.FE}{BD}=FA\hfill \end{array}$$
2. Now we need to look for a hint about side $FA$ . Looking at triangle $CAH$ we see that there is a line $FG$ intersecting it parallel to base $CH$ . This gives us another proportionality relation.
   $$\begin{array}{ccc}\hfill \frac{AG}{GH}& =& \frac{FA}{FC}(\mathrm{FG}\parallel \mathrm{CH}\mathrm{splits\; up\; lines}\mathrm{AH}\mathrm{and}\mathrm{AC}\mathrm{proportionally})\hfill \\ & \therefore & FA=\frac{FC.AG}{GH}\hfill \end{array}$$
3. We have 2 expressions for the side $FA$ .
   $$\begin{array}{ccc}& \therefore & \frac{FC.AG}{GH}=\frac{DC.FE}{BD}\hfill \end{array}$$