Calculation of the heights of any triangle

Question

thanks for helping me solving this problem

Let $ABC$ a triangle such : $AB=13$ ; $ AC=14$ ; $BC=15$

$(AI)$ and $(BJ)$ and $(CK)$ are the heights of the triangle $ABC$

1)Show that $BJ = 12$

2)Calculate $AI$ and $CK$

3)Propose a triangle whose sides and heights are natural numbers

My attemps:

1) I found the ratios of the similar triangles : $ICA$ and $ICB$ - $CKA$ and $JBA$ - $IAB$ and $KCB$ but i have always two unknowns .

2)I think that it's the same principle as the previous question

3) I have absolutely no idea

Thank you very much

Answer 1

You are correct that we have

$AB*CK = BC*AI = AC*BJ = 2*$Area of triangle.

$13*CK = 15*AI = 14*BJ$

This is three equations and three unknowns but they aren't independent. The can reduce to one unknown, say $BJ$ and $CK=\frac {14}{13}BJ$ and $AI=\frac {15}{13}BJ$.

But as $13^2 + 14^2 > 15^2; 13^2 +15^2 > 14^2; 14^2 +15^2 > 13^2$ the triangle has three accute angle and the perpendicular heights all intersect the sides. That is, $K$ is between $A$ and $B$ and $I$ is between $B$ and $C$ and $J$ is between $A$ and $C$.

Now consider the right triangle that formed:

$AK^2 + CK^2=AK^2 + (\frac {14}{13}BJ)^2 = AC^2 = 14^2$ And $BK^2 + CK^2=(13-AK)^2 + (\frac {14}{13}BJ)^2 = BC^2=15^2$

And so on.

You will have $7$ equations and $4$ unknowns.

Answer 2

Some hints:

1&2) I suggest just, for example, labelling $AJ=a, JC=b$. We then have: \begin{cases}a^2 + h^2 = 13^2 \\ b^2 + h^2 = 15^2 \\ a+b=14 \end{cases}

which can be easily solved, and then you can use the formula that @fleablood wrote to link the heights between themselves.

Although, in my opinion, the best way to solve this problem is using Heron's formula. Look it up!

3) Think of a type of triangle where heights are particularly nice to calculate...