geom walk through 40-40-100

1. Thought process - subdivision creates more interesting angles 2 possibilities 3x/4x or 4x/3x

2. First didn’t pan out despite isosceles and angle bisector.

3. Second created more ability to break everything down into combinations of two lengths.

4. Calc as many values as possible to prove the main triangle is isosceles

5. Went down the path of making a parallelogram on the right. Wasn’t useful but interesting.

6. Eventually found enough similar triangles to make an isosclese in the center and simplified a bit.

Alt:

• Per trig is possibility.

• You can translate the left hand subtriangle underneath the right hand one to make an isoscelese parallelogram! That makes its easy since it creates new parallel lines.

Constructions

1. Subdivide $\angle{BCD}$  such that $\angle{BCE} = 3x $   That makes $\triangle{ACE}$ an isosceles and AC = AE = BD.  This implies AD = BE.  

Let a = AD and BE  and  b = DE

1. Add a parallel segment DF to CE.  $\triangle{AFD}$ is similar to $\triangle{ACE} $  and also isosceles so  AF = a  and CF = b.

Similar Ratios

1. By similarity of $\triangle{ADF}$ and $ \triangle{ACE}$ , $ DF = CE \cdot \frac{a}{a+b} $  but triangles CFD and CEF are also similar implying  $ \frac{b}{DF}  = \frac{CE}{a} $

This simplifies to CE = $ \sqrt{(a +b)\cdot b} $

1. Now consider triangles CDE and CBD which are also similar.    So $ \frac{BD}{CD} = \frac{CD}{DE} $  or $ CD = \sqrt{BD \cdot DE} = \sqrt{(a+b) \cdot b}  $ which is the same as CE above.

So CDE is an isoscleses triangle and  and triangle ADC and BCE are congruent.   That means 4x = 180 - 14x  or x = 10 and    $ \angle{CAB} = 40$