How to calculate the length $AH$?

Question

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?

Answer 1

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 $\Delta ABE$.

Answer 2

Here is a solution.

I had a small calculation mistake in the previous solution which I posted.

Answer 3

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.

Answer 4

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:

1. Side $AB$.

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

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

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

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

6. Line $\mathcal{L}_{BD}=BD$

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

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

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

10. 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}