How is this quadratic derived?

Question

I'm reading 'Geometry by Roger Fenn' (https://www.google.co.uk/books/edition/Geometry/b1HlBwAAQBAJ?hl=en&gbpv=1&printsec=frontcover) and on Page 19 and 20 he goes into deriving Pi using Geometric means. I don't get how he derived the quadratic, I will copy the text from the book, and screenshot image of the triangles. Link above has actual pages available for reference.

Given the following:

The circle has diameter of 1.

Area of Triangle $$ADB = \frac{1}{2}AD \cdot BD = \frac{1}{2}AB \cdot CD = \frac{1}{2} CD$$

By the above equation the triangles ADB and ADC are equal, but they are clearly different size, does this make sense to anyone?

By Pythagoras' Theorem $$AD^2 = AB^2 - BD^2 = 1 - BD^2$$

Writing $$CD = \frac{DE}{2}$$ and eliminating the length AD we obtain the following quadratic equation for BD^2

$$4BD^4 - 4BD^2 + DE^2 = 0$$

I can see that if you combine Pythagoras' equation for both ADB and ADC you can get $$4BD^2 + DE^2$$ by substitution and re-arranging, however I do not know where the first term comes from? Can anyone point me in the right direction? There must be a way to derive this that I'm not seeing.

Answer 1

You say:

By the above equation the triangles ADB and ADC are equal,

That's not what that equation says. It finds the area of right triangle ADB in two ways. One that uses its two legs (AD and DB) and one that uses the hypotenuse and an altitude (AB and CD).

Now if $$AD^2=1-BD^2$$ and you multiply through by $BD^2$:

$$AD^2BD^2=BD^2-BD^4$$

and use the relation that $AD\,BD=CD$ (as established by the part where you mention triangles ABD and ADC):

$$CD^2=BD^2-BD^4$$

and multiply through by $4$:

$$4CD^2=4BD^2-4BD^4$$

and use that last relation that $2CD=DE$:

$$DE^2=4BD^2-4BD^4$$

and this is the relation that the author arrives at.