# Decipher Triangle Theorem in David Wells Book

It's now about the tenth time I've tried to read David Wells's book "You are a Mathematician" and not got past the first chapter because I can't understand his writing.

There is a theorem about triangles which I'm sure is very simple expressed in these two images:

However, the relationship between the first diagram and the second is not made clear (1.4 and 1.6), and there are no labels on the second one to help with the explanation.

He seems to be saying that in a triangle cut into two parts by a line from one vertex to the opposite side, if the angle sums of the three triangles thus formed are equal then the angle sums of the smaller triangles must be equal to the sum of the angles on either side of the point where the line form the vertex meets the opposite side.

However I see no argument explaining why this should be so.

Could someone please provide a clearer explanation, perhaps with the aid of a diagram?

Also could someone please give me an idea of how clear they consider the explanation given in the book to be? I'm curious to know if my confusion is "normal."

• A weird title for a book ! As long as the picture cannot be seen directly (hope someone edits the question to change this) , try to describe what the author wants to prove. It is somehing about angle sums ? – Peter Feb 24 at 12:17
• I only see one image when I click the link. – Jens Feb 24 at 12:28
• I am surprised that you insist on reading a book you don't go along with its writer's style. There is no shortage of excellent geometry books! – NoChance Feb 24 at 12:41
• @peter doesn't "He seems to be saying that in a triangle...." do just that? – Robin Feb 24 at 13:09
• @Robin Well possible that the cumbersome formulated statement tells it. – Peter Feb 24 at 13:12

Suppose the sum of the angles of any triangle is some constant $$\pi$$. The sum of the angles of the large triangle is the sum of the angles of the two small triangles, minus $$X$$ and $$Y$$. Hence: $$\pi=\pi+\pi-X-Y, \quad\text{that is:}\quad X+Y=\pi.$$
The overall argument from Fig. 1.4 and Fig. 1.6 appears to be as you suggest, and the argument from Fig. 1.6 in particular appears to be that if, as conjectured, the angle sum of any triangle has a constant value $$S$$, then, in the figure below: \begin{align*} 2S & = (\text{angle sum of } \triangle ABD) + (\text{angle sum of } \triangle ADC) \\ & = (\alpha_2 + \beta + X) + (\alpha_1 + Y + \gamma) \\ & = ((\alpha_1 + \alpha_2) + \beta + \gamma) + (X + Y) \\ & = (\text{angle sum of } \triangle ABC) + (X + Y) \\ & = S + (X + Y), \end{align*} whence $$S = X + Y$$. I don't know why he expresses this conclusion as [emphasis mine] "the angle sum of each of the smaller triangles must be $$X + Y$$": that does seem rather confusing. Perhaps it is intended to express the fact that $$S$$ is subtracted from both sides of the equation, leaving the angle sum of one small triangle?
I rather agree. I think Wells's argument from Fig. 1.4 is that when a line rotates, the angle it makes with any other (fixed) line changes by the same amount, viz. the angle of the rotation. Rather more clearly, he concludes from this that the angle sum of $$\triangle ABC$$ in Fig. 1.4 [no relation to $$\triangle ABC$$ in my annotated version of Fig. 1.6 - I'm sorry about the clash of notation] is equal to the angle sum of $$\triangle AB'C'$$.