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. 
 A: 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.
A: First Solution:
It's known that a triangle is right angled if and only if the three edges satisfies the Pythagorean theorem (this statement follows from the Low of Cosines). So we have only to verify the Pythagorean theorem for the triangle. 
Named $a=BC$ we have:
\begin{gather}
a^2+b^2 = h_c^2 + a_1^2 + b_1 c = a_1b_1 + a_1^2 + b_1c=\\
a_1(b_1+a_1) + b_1 c = a_1c+b_1c=(a_1+b_1)c = c^2 
\end{gather}
Where the first equality follows from Pythagorean theorem $a^2=h_c^2+a_1^2$, and we have used that $a_1+b_1=c$ because they are the projections.

Second solution:
Take a cartesian plane and put the triangle $ABC$ in it: $A=(-a_1,0)$, $B=(b_1,0)$, $C=(0,h_c)$.
Now take the line $AC$ and $CB$. Their angular coefficients are rispectively
\begin{gather}
AC:\quad \frac{h_c-0}{0-(-a_1)} = \frac{h_c}{a_1}\\
CB:\quad \frac{h_c-0}{0-b_1} = -\frac{h_c}{b_1}
\end{gather}
Multipling now the angular coefficients we obtain:
$$
\frac{h_c}{a_1} \cdot \left( - \frac{h_c}{b_1}\right) = \frac{h_c^2}{a_1b_1} = -1
$$
So the two lines are perpendicular and the angle in $C$ is a right angle.
Edit: I realize now that these arguments work for the angle in $A$ and $B$ acute.
A: If you have that $h_c^2=a_1b_1, a^2=ca_1$ and $b^2=cb_1$ then $a^2b^2=c^2h_c^2=4A_{\triangle ABC}^2$. Thus, $A_{\triangle ABC}=\frac{ab}{2}$ and $a$ is perpendicular to $b$.
