# Ellipse inscribed on a quadrilateral

The problem is:

Given that an ellipse is inscribed on a convex quadrilateral and each one of it's diagonals pass through one foci of the ellipse show that the product of the opposite sides are equal.

Now I've been thinking about this one for a while and I could transform this in an angle problem, please if someone can show that the two marked angles on the pic are equal I will appreciate (A and B are the focis of the ellipse $EHLG$ is the quadrilateral):

If at the end of this 6 days nobody comes with a plane geometry answer I'll reward the trigonometric solution

• Am I allowed to post a shorter answer? – najayaz Mar 18 '15 at 13:07
• sure you are, if it is trigonometric though you will still need to wait a little more to get the 50 points – onlyme Mar 18 '15 at 13:21

This is a complete proof, and it might get a bit repetitive because I'm using the Sine Law again and again. I'm trying to find something more elegant, but this is the only thing thing I could come up with right now.( I've used a different diagram because my internet wasn't working while I was working this out. Just skip over to the ending note if you just want to see how the angles are equal.)

$DEGF$ is the quadrilateral and $A$ and $B$ are the foci.

Using sine law in $\Delta DBE$ you get $\dfrac{DE}{BE}=\dfrac{\sin\angle DBE}{\sin\angle BDE}$ and in $\Delta GBE$ you get $\dfrac{GE}{BE}=\dfrac{\sin\angle GBE}{\sin\angle BGE}$. Dividing the two you'll obtain: $\dfrac{DE}{GE}=\dfrac{\sin\angle DBE}{\sin\angle BDE}\cdot\dfrac{\sin\angle BGE}{\sin\angle GBE}\tag{i}$

Using sine law in $\Delta BDF$ you get $\dfrac{FD}{FB}=\dfrac{\sin\angle DBF}{\sin\angle BDF}$ and in $\Delta BGF$ you get $\dfrac{FG}{FB}=\dfrac{\sin\angle GBF}{\sin\angle BGF}$. Dividing the two you'll obtain: $\dfrac{FD}{FG}=\dfrac{\sin\angle DBF}{\sin\angle BDF}\cdot\dfrac{\sin\angle BGF}{\sin\angle GBF}\tag{ii}$

Dividing (i) and (ii):$\dfrac{DE\cdot FG}{GE\cdot FD}=\dfrac{\sin\angle BGE}{\sin\angle BDE}\cdot\dfrac{\sin\angle BDF}{\sin\angle BGF}\tag{iii}$

Using sine law in $\Delta DFI$ you get $\dfrac{DI}{FI}=\dfrac{\sin\angle DFE}{\sin\angle FDG}$ and in $\Delta GFI$ you get $\dfrac{GI}{FI}=\dfrac{\sin\angle GFE}{\sin\angle FGD}$. Dividing the two you'll obtain: $\dfrac{DI}{GI}=\dfrac{\sin\angle DFE}{\sin\angle FDG}\cdot\dfrac{\sin\angle FGD}{\sin\angle GFE}\tag{iv}$

Using sine law in $\Delta DEI$ you get $\dfrac{DI}{EI}=\dfrac{\sin\angle DEF}{\sin\angle EDG}$ and in $\Delta GEI$ you get $\dfrac{GI}{EI}=\dfrac{\sin\angle GEF}{\sin\angle EGD}$. Dividing the two you'll obtain: $\dfrac{DI}{GI}=\dfrac{\sin\angle DEF}{\sin\angle EDG}\cdot\dfrac{\sin\angle EGD}{\sin\angle GEF}\tag{v}$

Combining (iv) and (v): $\dfrac{DI}{GI}=\dfrac{\sin\angle DFE}{\sin\angle FDG}\cdot\dfrac{\sin\angle FGD}{\sin\angle GFE}=\dfrac{\sin\angle DEF}{\sin\angle EDG}\cdot\dfrac{\sin\angle EGD}{\sin\angle GEF}\tag{vi}$

Doing the same thing with $DB$ and $GB$ you will get $$\dfrac{DB}{GB}=\dfrac{\sin\angle DFE}{\sin\angle GFE}\dfrac{\sin\angle BGF}{\sin\angle BDF}=\dfrac{\sin\angle DEF}{\sin\angle GEF}\dfrac{\sin\angle BGE}{\sin\angle BDE}\tag{vii}$$

Dividing (vi) and (vii) you get:$$\dfrac{\sin\angle FGD}{\sin\angle FDG}\cdot\dfrac{\sin\angle BDF}{\sin\angle BGF}=\dfrac{\sin\angle DGE}{\sin\angle GDE}\cdot\dfrac{\sin\angle BDE}{\sin\angle BGE}\tag{viii}$$

Now due to the Property 1 on this link, $\angle FGD=\angle BGE, \angle GDE=\angle BDF,\angle DGE=\angle BGF$ and $\angle BDE=\angle FDG$. Making the necessary substitutions, relation (viii) becomes:$$\sin^2\angle BGE \cdot\sin^2\angle BDF=\sin^2\angle BDE \cdot\sin^2\angle BGF$$ $$\sin\angle BGE \cdot\sin\angle BDF=\sin\angle BDE \cdot\sin\angle BGF\tag{ix}$$

Combining (iii) and (ix): $DE\cdot FG=GE\cdot FD$

NOTE: Making use of the relations (vi),(vii) and (ix) it is easy to see that $\dfrac{DI}{GI}=\dfrac{DB}{GB}$. By the Angle Bisector Theorem, this means that $\angle DBF = \angle GBF$ and similarly $\angle FAG = \angle EAG$.