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. 

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.
