Given:

The line passing through E and D is tangent to the circle at D.

AC = AB

Take a look at the drawing.

Let's see if geogebra and experimental math can provide an answer.

1) Hide the xy axis.

2) Input point (1,1), it will be called A.

3) Use the circle with center through point to create a circle with center at A. When the circle is created point B will be also.

4) Draw a line through points A and B.

5) Draw a perpendicular line to that line through A.

6) Use the intersection tool to findd the points of intersection of the circle and the perpendicular line. Points C and D will be created.

7) For cosmetic purposes rename C to Z and D to C. Then hide Z.

8) Create poly1 using the polygon tool and clicking ABCA in that order.

Hide the circle and the horizontal and vertical line. You should just have triangle ABC on the screen.

This triangle is very powerful. Its power comes from the fact that it is dynamic. I can pull on point A or B to create any isosceles triangle I want. Try it, keeping a close eye on a1, b1 and c1 ( these are the lengths of the sides of the triangle) in the segment pane.

9) Use the circle through a point tool and click on C and drag the circle till it touches A.

10) Use the intersect tool to find the points of intersection of the triangle and the new circle. D will be created on CB.

11) Draw a tangent to the circle through D.

12) Use the intersection tool to find the intersection of this tangent and AB. It will be called E.

13) Use the polygon tool and click EBDE to create poly2.

14) Enter in the input bar ratio = poly1 / poly2. The ratio of the two triangles will appear in the algebra pane underneath number.

15) Adjust rounding in options to 15 decimal places. The finished construction should look like the picture below.

16) Now here is the whole point of this. Take and drag A or B around the screen watching the variable ratio closely. If you have done this right it will remain at

ratio = 5.828427124746185

No matter what isosceles triangle ABC you create the ratio of ABC's area to triangle EBD's area remains constant.

Are we done? Not yet. The hallmark and the backbone of experimental mathematics are the LLL and PSLQ algorithms. We will use them now to get a closed form for 5.828427124746185.

There are several ways to get what we need. Today we will use wolfram alpha. Enter:

rootapproximant(5.828427124746185)

you will be pleasantly surprised with

We can conjecture that the ratio of the areas of ΔEBD to ΔEBD is

We are done!

]]>