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  1. I started with a $40^\circ$ angle:

enter image description here

  1. I then copied that to the other side with a compass, I then added a perpendicular line on each side to make it a right triangle, making sure both are the same length:

enter image description here

  1. I then drew a triangle with the width of the other two triangles, and with a height double that of the other triangles:

enter image description here

  1. Then, I drew a fourth triangle with the height of the small and large triangle combined, and the length of three times the small triangle:

enter image description here

  1. Then, I drew an arc on a circle with a radius of the width of the two small triangles, and it intersected the largest triangle:

enter image description here

  1. I copy the triangle to the right:

enter image description here

  1. I draw a segment from the intersection of the arc and the largest triangle to the furthest right point of the furthest triangle:

enter image description here

  1. But then, the angle from the base and the segment just drawn is $~9^\circ$, and the other is $~30^\circ$:

enter image description here

I'm confused as to why they aren't equal, as the largest triangle is being split into 2 parts, one is 1/3 of it, the other is 2/3

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    $\begingroup$ Welcome to Math.SE! ... The upshot of your question is (referencing your final figure): In isosceles $\triangle QJK$ with $O$ on $\overline{JK}$ such that $|OK|=\frac13|JK|$, why isn't $\angle OQK=\frac13\angle JQK$? The short answer is that trisecting lengths and trisecting angles almost-never agree like that. (There's one magic angle (about $62.1^\circ$) where this strategy does work.) We shouldn't really expect this to happen, anyway. After all, a median of a triangle —which joins a vertex to the midpoint of the opposite side, bisecting that side— typically doesn't bisect an angle. $\endgroup$
    – Blue
    Apr 21 at 5:32

1 Answer 1

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Please refer to my simplified diagram. enter image description here

Hence from my diagram, by properties of parallel lines,$$\angle ABC=40°$$ $$BE=4+2+2+2=10 \ \text{units}$$ $$ED=2\tan 40°$$ Hence $$\tan \angle EBD =\frac{ED}{BE}=\frac{2\tan 40°}{10}$$ $$\implies \angle EBD=\arctan\bigg(\frac{\tan 40°}{5}\bigg)=9.5266°$$ Thus $\angle ABD=40°-9.5266°=30.4733°$.

You're not essentially trisecting the angle. You're actually trisecting the side $AB$. If you measure it, $CD=2\cos 40°,AD=4\cos 40°$.

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    $\begingroup$ To anyone wondering why this paper is so crooked, I drew this diagram on the back of a question paper, and I didn't have a ruler. I used my comb. Hence the diagram is not to scale :) $\endgroup$
    – Gwen
    Apr 21 at 14:13

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