# Show that the angles satisfy $x+y=z$

How can I show that $x+y=z$ in the figure without using trigonometry? I have tried to solve it with analytic geometry, but it doesn't work out for me. • Are all figures in the image intended to be squares? – Edward Jiang Jan 9 '16 at 20:00
• Yes! I forgot to mention that. – Hamid Mohammad Jan 9 '16 at 20:01
• See the solution detailed here. It also comes with a nice extension problem which can be approached similarly. – πr8 Jan 9 '16 at 20:10 (This space intentionally left blank.)

Hint: denoting the points like on the picture below, triangles $EGD$ and $DGF$ are similar (why?). • Very neat indeed. – Lubin Jan 9 '16 at 20:40
• Please could you explain how these triangles are similar? – Fly by Night Jan 9 '16 at 22:46
• @FlybyNight: Compare the ratios of corresponding sides of $\angle G$ in both triangles. (This is effectively the product relation provided in Batominovski's answer.) – Blue Jan 9 '16 at 23:53
• @Blue Hi Blue. My comment was aimed at wojowu, in the hope that s/he might add the detail to his/her original answer. – Fly by Night Jan 10 '16 at 0:13

Let $X$, $Y$, and $Z$ be the apexes of the angles $x$, $y$, and $z$, respectively. Also, let $P$ be the common intersection of the red lines. Show that $ZP^2=ZX\cdot ZY$. Thus, the circumcircle of the triangle $PXY$ is tangent to $PZ$ at $P$. This will prove that $\angle YPZ=\angle YXP=x$.

Equivalently, you have to prove that $$\underbrace{\mathrm{arctan}(1/3)}_{x}+\underbrace{\mathrm{arctan}(1/2)}_{y}=\underbrace{\mathrm{arctan}(1)}_{z}.$$ This is pretty clear by addition of tangents, indeed $$\tan(x+y)=\frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}=1 \implies x+y=\frac{\pi}{4}.$$

• The OP clearly states he wanted a non-trigonometric solution. – Edward Jiang Jan 9 '16 at 20:09

Imagine that all the the squares are 1 by 1 and so the rectangle has base 3 and a height of $1$.

There are three right-angled triangles in the diagram. The one with angle $x$ has base 3, and height 1. The triangle with angle $y$ has base 2 and height 1. The triangle with angle $z$ has base 1 and height 1.

Using the standard trig' ratio $\tan \theta = \frac{\mathrm{opp}}{\mathrm{adj}}$, we get $\tan x = \frac{1}{3}$, $\tan y = \frac{1}{2}$ and $\tan z= \frac{1}{1}=1$.

There is a well-know formula for angle addition:

$$\tan(\alpha+\beta) = \frac{\tan \alpha + \tan \beta}{1-\tan \alpha \tan \beta}$$

Applying this formula to the case of $\alpha =x$ and $\beta = y$ gives: $$\tan(x+y) = \frac{\tan x + \tan y}{1-\tan x \tan y}=\frac{\frac{1}{3}+\frac{1}{2}}{1-\frac{1}{3}\cdot\frac{1}{2}}=1$$ It follows that $\tan(x+y)=\tan z$. Since $0^{\circ} < x<y<z < 90^{\circ}$ it follows that $$\tan(x+y) = \tan z \iff x+y = z$$

• "... without using trigonometry?" – Blue Jan 9 '16 at 22:36
• @Blue All of the answers given so far involve trigonometry to some degree, even yours. The fact that you don't mention sine, cosine or tangent is irrelevant. The answer to the OP's question should be: no such answer exists. Trigonometry is the science of triangles using lengths, angles and their ratios. I could re-phrase my answer in terms of lengths and proprtion, but that's really just trigonometry, but without the nice simple place-holders of sine, cosine and tangent. – Fly by Night Jan 9 '16 at 22:40
• @fleablood You're absolutely right. I hadn't read the "without trigonometry" until after I'd written my answer. However, being asked to solve this problem without trigonometry is, like I said, being asked to solve it without using angles, lengths, and their ratios. I'll leave my answer here because I think it might help people who Google this question in the future, and who are willing to use trigonometry to solve problems involving triangles. – Fly by Night Jan 9 '16 at 22:55
• @Blue I agree. Your solution is great! – Fly by Night Jan 9 '16 at 23:42
• @fleablood Be careful with the accessibility statement. We teach children trigonometry from the ages of 13/14 in the UK. The angle addition formulae come along at around 16/17. My answer is perfectly accessible for someone able to generate the graphic like the OP did. It's not like I'm using cohomology. – Fly by Night Jan 9 '16 at 23:45