A geometry problem - easy with trigonometry, harder without it

Question

Consider a $\triangle ABC$, with $\angle A=15^{\circ}, \angle B=55^{\circ}, \angle C=110^{\circ}$.

Prove that $c^2=ab+b^2$.

This is from my math teacher. I solved it in 10 minutes with trigonomery, not that difficult. My question is: it is possible to prove it without trigonometry? Please help! I am thankful for every solution!

Well, every time you deal with triangles, you're doing trigonometry, whether it's explicitly expressed or not. So what you ask for is impossible, strictly speaking. However, one may relate various properties of the triangle together to solve this problem without directly mentioning the trigonometric relationships as we know them. — Allawonder, May 21 '18 at 21:59
That is you need to prove that $ c^2=a^2+b^2-2ab \cos C = ab+b^2 ; i.e., a/b = \sin 15^{\circ}/ \sin 55^{\circ}= 1- 2 \cos 70 ^{\circ}$.Can take it further? — Narasimham, Jun 17 '18 at 07:12

score 7 · Accepted Answer · edited May 21 '18 at 21:52

7

Let $D$ be the point where the bisector of angle $C$ hits side $AB$.

Then we get

$\angle BCD=55^\circ$$\\[4pt]$
$\angle BDC=70^\circ$$\\[4pt]$
$\angle ACD=55^\circ$$\\[4pt]$
$\angle ADC=110^\circ$

hence $\Delta BDC$ is isosceles, with $|DB|=|DC|$, and $\Delta ABC$ is similar to $\Delta ACD$.

The factor for scaling side lengths of $\Delta ABC$ to corresponding side lengths of $\Delta ACD$ is $$\frac{|AC|}{|AB|}=\frac{b}{c}$$ hence, we get \begin{align*} |DC|&=\left(\frac{b}{c}\right)a\\[4pt] |AD|&=\left(\frac{b}{c}\right)b\\[4pt] \end{align*} But then also \begin{align*} |AD|&=c-|DB|\\[4pt] &=c-|DC|\\[4pt] &=c-\left(\frac{b}{c}\right)a\\[4pt] \end{align*} so \begin{align*} &c-\left(\frac{b}{c}\right)a=\left(\frac{b}{c}\right)b\\[4pt] \implies\;&c^2-ab=b^2\\[4pt] \implies\;&c^2=ab+b^2 \end{align*}

edited May 21 '18 at 21:52

String

18,395

answered May 21 '18 at 21:28

quasi

58,772

Is it OK that I added a figure summarizing your fine work? Otherwise please tell me or just delete it! – String May 21 '18 at 21:53
@String: Looks good! Thanks! – quasi May 21 '18 at 21:55
Nice! Also feel free to move it around if it fits better earlier in your post. – String May 21 '18 at 21:57

score 5 · Answer 2 · answered May 21 '18 at 23:45

5

All that matters is that $\angle C = 2 \angle B$:

$$\triangle ABC \sim \triangle AB^\prime B \qquad\to\qquad \frac{c}{b}=\frac{a+b}{c}$$

answered May 21 '18 at 23:45

Blue

75,673

A geometry problem - easy with trigonometry, harder without it

2 Answers2