How to calculate the length $AH$? We have a triangle $ABC$ and the heights $AD$ and $BE$ that intersect in $H$.
The following length are given: $$AB=12, \ BD=4, EC=  8, \ AE=6$$ How can we calculate $AH$ ?
Does the orthocenter divide a length by a specific ratio?
 A: To prove: $AH=2R\cos(A)$ (symbols have their usual meanings in a triangle)
Follow $$AH = \dfrac{AE}{\cos(\angle HAE)} \ \left(from \ right-angled \ \Delta HAE\right)\\= \dfrac{AE}{\cos(\angle DAC)}=\dfrac{AE}{\cos(90^{\circ}-C)}=\dfrac{AE}{\sin(C)}$$
and $AE=AB\cos(A)$ (from right angled $\Delta ABE$)
so $$AH = \dfrac{AB\cos(A)}{\sin(C)} = \dfrac{c}{\sin(C)}\cos(A) = 2R\cos(A) \ \left(by \ sine \ rule\right)$$
Knowing $AH = 2R\cos(A) = \dfrac{b}{\sin(B)}\cos(A)$ (by sine rule)
Find $b=CA=CE+EA=14$, $\sin(B)=\dfrac{AD}{AB}=\dfrac{8\sqrt{2}}{12}=\dfrac{2\sqrt{2}}3$ from the right angled $\Delta ABD$ and $\cos(A)=\dfrac{AE}{AB}=\dfrac12$ from the right angled $\Delta ABE$.
A: Construction of $\triangle ABC$
based on given data is impossible,
since condition $|EC|=8,\ |AE|=6$
contradicts the other data in this construction.

However, triangle $ABC$ can be constructed
given three lengths
$|AB|=12=c,\ |BD|=4,\ |AC|=8+6=14=b$
as follows:

*

*Side $AB$.


*Circle $\Omega_B(B,|BD|)$.


*Point $C_m=(A+B)/2$.


*Circle $\Omega_{C_m}(C_m,\tfrac12\,|AB|)$.


*Point $D=\Omega_{C_m} \cap \Omega_B$.


*Line $\mathcal{L}_{BD}=BD$


*Circle $\Omega_A(A,|AC|)$.


*Point $C=\Omega_A \cap \mathcal{L}_{BD}$.


*Point $E_1=\Omega_{C_m} \cap \mathcal{L}_{AC}$.


*Point $H=\mathcal{L}_{AD}\cap \mathcal{L}_{BE_1}$
The foot of the other height is named $E_1$ fo avoid confusion
with the contradictory point $E$, mentioned in the OP.
\begin{align}
\triangle ABD:\quad
|AD|&=
\sqrt{|AB|^2-|BD|^2}
=8\sqrt2
,\\
\triangle ADC:\quad
|CD|&=
\sqrt{|AC|^2-|AD|^2}
=2\sqrt{17}
,\\
|BC|=a&=|BD|+|CD|=
4+2\sqrt{17}
.
\end{align}
Given $a,b,c$, we can find
the area $S$ and the circumradius $R$ of $\triangle ABC$
as
\begin{align}
S&=\tfrac12\,|AD|\cdot|BC|
=4\sqrt2\,(4+2\sqrt{17})
,\\
R&=\frac{abc}{4S}
=\frac{21\sqrt2}4
.
\end{align}
Finally, using that in any triangle
$ABC$ these relations hold true:
\begin{align}
a^2+|AH|&=b^2+|BH|^2=c^2+|CH|^2
=4R^2
,
\end{align}
the answer therefore is
\begin{align}
|AH|&=
4R^2-a^2
=\tfrac{\sqrt2}2\,(16-\sqrt{17})
\approx 8.398
.
\end{align}
Note that in this construction
the value of the distance from $C$
to the foot of the height from $B$
is clearly dictated
by the power of the point $C$
wrt $\Omega_{C_m}$
\begin{align}
|CE_1|&=\frac{|CD|\cdot|BC|}{|AC|}
=\tfrac27\,(17+2\sqrt{17})
\approx7.2132
\mathbf{\ \ne 8=|EC|}
,\quad\text{given in OP}
.
\end{align}
A: Here is a solution.
I had a small calculation mistake in the previous solution which I posted.

A: I've made a scaled sketch of the problem in Geogebra by assuming the segments $AB, \, AE$ and $CE$ to be true and then finding the length of segment $BD$ for cross validation. Two such triangles are possible and I think there might be some error in your values else the answer is "No such triangle is possible". Also, let me know if I'm wrong.


