For reference:
In triangle $ABC$, the ratio of the squares of sides $AC$ and $AB$ is equal to the ratio of the projections of these sides on side $BC$. Calculate $\angle ABC$, if $\angle BCA = 80^\circ$ (Answer:$10^\circ$)
My progress:
I drew the picture but could not see the similarity with the given data.
Is it possible to determine similarity with this data alone, or is there a lack of data?
$\dfrac{AC^2}{AB^2} =\dfrac{m}{n}$
With a hint from Math:
$\dfrac{AC^2}{AB^2} =\dfrac{m}{n}$
$AD \perp BC \implies \\ \triangle ABD: AB^2 = AD^2+n^2\\ \triangle ACD: AC^2=AD^2+m^2\\ \frac{AD^2+m^2}{AD^2+n^2}=\frac{m}{n}\rightarrow nAD^2+nm^2=mAD^2+mn^2\rightarrow \\AD^2(n-m)=mn(n-m)\implies AD^2 = mn$
Therefore we have a metric relation of the right triangle and therefore $\angle A = 90^o$ and $\angle B = 10^o$