In square $c^2=a^2+2b^2$, Find the angle $\alpha$

Question

In square in the image $c^2=a^2+2b^2$. Find $\alpha$.

I tried to solve it using law of cosines. m is the side of the square:

$m^2=b^2+a^2-2abcos(\alpha)$

And if the angle of intersection of c and b is $\beta$:

$m^2=b^2+c^2-2bccos(\beta)$

And as the other angle is $360-(\alpha+\beta)$ and $cos(360-(\alpha+\beta))=cos(\alpha+\beta)$ and diagonal of square is $m\sqrt{2}$ :

$2m^2=c^2+a^2-2accos(\alpha+\beta)$

If we subtract

$m^2=b^2+c^2-2bccos(\beta)$

$m^2=b^2+a^2-2abcos(\alpha)$

We get:

$0=c^2-a^2+2abcos(\alpha)-2bccos(\beta)$ as $c^2-a^2=2b^2$ we get $2bccos(\beta) -2abcos(\alpha)=2b^2$ $\to$ $b=ccos(\beta)-acos(\alpha)$

Then we equalize RHS of two below and simplify:

$2m^2=2b^2+2a^2-4abcos(\alpha)$

$2m^2=c^2+a^2-2accos(\alpha+\beta)$

We get:

$2bcos(\alpha)= ccos(\alpha+\beta)$ $\to$ $b=\frac{ccos(\alpha+\beta)}{2cos(\alpha)}$

And if we equalize RHS of two below and simplify:

$b=ccos(\beta)-acos(\alpha)$

$b=\frac{ ccos(\alpha+\beta)}{2cos(\alpha)}$

We get:

$2ccos(\alpha)cos(\beta)-2acos^2(\alpha)=ccos(\alpha)cos(\beta)- csin(\alpha)sin(\beta)$ $\to$ $ccos(\alpha-\beta)= 2acos^2(\alpha)$

But then I can't go any further.Can you finish my way or can you suggest maybe easier and better solution?

Answer 1

I propose a simpler solution. As shown in the figure above, construct $\triangle AFD\cong\triangle AEB$. Since $\angle DAF=\angle BAE$, we have $$\angle EAF=\angle EAD+\angle DAF=\angle EAD+\angle BAE=\frac{\pi}2\,.$$ With $EA=FA=b$ we obtain $EF=\sqrt{2}b$. We then observe $\triangle DEF$. Note that $$DF^2 = c^2=a^2+(\sqrt{2}b)^2=DE^2+EF^2\,.$$ Hence $\angle DEF=\frac{\pi}{2}$, and we conclude that $$\alpha=\angle AEF+\angle DEF=\frac{\pi}4+\frac{\pi}2=\frac{3\pi}4\,.$$

Answer 2

From the figure above, we have $s=\sqrt{b^2-u^2}, t=\sqrt{a^2-u^2}$ and $v^2-u^2=a^2+b^2$ since $c^2=a^2+2b^2.$

On the other hand, $s+t=u+v.$ So, $$v^2-u^2=a^2+b^2$$ $$(s+t-u)^2-u^2=a^2+b^2$$ $$(s+t)^2-2u(s+t)=a^2+b^2$$ $$a^2-u^2+b^2-u^2+2st-2u(s+t)=a^2+b^2$$ $$st-u^2=u(s+t)$$ On the other hand, $$\tan(x+y)=\frac{\tan x+\tan y}{1-\tan x\tan y}=\frac{u(s+t)}{u^2-st}=-1.$$ So, $\alpha=x+y=\arctan(-1)=135^{\circ}.$