Necessary and sufficient conditions for a right triangle 
Show that if for $\triangle ABC$ the equalities $h_c^2=a_1b_1$ and $b^2=b_1c$ are true where $h_c$ is the height, $AC=b, AB=c$ and $a_1$ and $b_1$ are the the projections of $BC$ and $AC$ on $AB$, then the triangle is right angled.

I want to demonstrate that together the conditions are sufficient without using the Law of Cosines or the Pythagorean theorem. The relations in a right triangle that we have studied and can use in the problem: $h_c^2=a_1b_1, a^2=ca_1$ and $b^2=cb_1$ where $BC=a, AC=b, AB=c$ and $a_1$ and $b_1$ are the projections of $AC$ and $BC$ on $AB$. 
Counterexample:

$\triangle ABC$ $(AC<BC)$ is right triangle and then $h^2=a_1b_1$. Let $M$ lie on $BH$ and $HM=AH=b_1$. For $\triangle MBC$ $h^2=a_1b_1$ but it is NOT A RIGHT TRIANGLE. 

I think your example gives the key for proving the statement. 
Indeed if one assumes that $H $ is an internal point of $AB$ the proof that $\angle ACB$ is right angle trivially follows from $h_c^2=a_1b_1$ in view of similarity of the right triangles $AHC$ and $CHB$ and the fact that two complimentary angles add at the vertex $C$. 
Thus the triangle $ABC$ can be not right-angled only if $H$ is not an internal point of $AB$. This is however impossible due to the condition $b^2=b_1c$. To prove this assume that the point $H $ is not internal. Let $A'$ be the image of $A$ upon reflection over the altitude $CH$. Since the point $H$ is an internal point of the segment $A'B $, $\angle A'CB $ is right angle, as was shown above. Therefore:
$$A'C^2=A'H\cdot A'B\implies 
b^2=b_1(a_1+b_1).$$
Together with the assumption 
$$b^2=b_1c=b_1|a_1-b_1|$$ 
this however implies that at least one of $a_1,b_1$ is $0$ which contradicts to the assumption $h_c^2=a_1b_1$.
Observe that it is possible to perform the above proof also in the opposite order (i.e. start with the assumption $b^2=b_1c$ and continue with $c^2_h=a_1b_1$). The common idea is: if $H$ is an internal point of $AB$ any of the conditions $a^2=a_1c$, $b^2=b_1c$, $c^2_h=a_1b_1$ alone is sufficient for the proof. And if any two of the three conditions are satisfied the point $H$ have to be internal.
