# Math Help - triangle + equations

1. ## triangle + equations

This is my problem: We have a triangle ABC, that has a right angle at the vertex C. First we draw the height v from the vertex C, the heigh v intersects the line AB in point D. Now we draw 2 lines from point D, x and y, one that is parallel with line AC, the other parallel with line CB. Now I have to prove that c*x*y=v^3
this is the photo:

I know the problem is relatively simple: (using the pythagorean theorem multiple times )it just requires equation solving but I'm really lazy/stupid so if I could get any hints or help that would be really great...

2. Hello, snezic!

If we're allowed to use trig, I have a solution.
(And I'm sure someone will find a shorter or neater one.)

We have a triangle $ABC$, that has a right angle at the vertex $C.$
We draw the altitude $v = CD.$
Now we draw two lines from point $D$:
. . $x = DE$ parallel to $AC,\;\text{ and }\;y = DF$ parallel to $BC.$

Prove that: . $cxy\:=\:v^3$

Code:
    B o
|  *
|     *
|     θ  *   D
E o - - - - - o
|     x    /:  *
|         / :     *
|        /  :        *
|       / θ :           *
|      /    :              *
y|     /v    :y                *
|    /      :                    *
| θ /       :                       *
|  /        :                          *
| /         :                             *
|/          :                             θ  *
C o - - - - - o - - - - - - - - - - - - - - - - - o A
x     F

Note that all the right triangles are similar.

$\begin{array}{ccccccccc}\text{In }\Delta DFA\!: &\csc\theta &=& \dfrac{DA}{y} &\Rightarrow& DA &=& y\csc\theta & [1] \\ \\[-3mm]
\text{In }\Delta BED\!: & \sec\theta &=&\dfrac{BD}{x} & \Rightarrow& BD &=& x\sec\theta & [2] \end{array}$

. . $\begin{array}{cccccc}\text{In }\Delta DFC\!: & \csc\theta &=& \dfrac{v}{x} & [3] \\ \\[-3mm]
\text{In }\Delta DEC\!: & \sec\theta &=&\dfrac{v}{y} & [4] \end{array}$

Substitute [3] and [4] into [1] and [2]: . $\begin{array}{cccccc}DA &=& y\cdot \dfrac{v}{x} &=& \dfrac{vy}{x} & [5]\\ \\[-3mm]
BD &=& x\cdot\dfrac{v}{y} &=& \dfrac{vx}{y} & [6] \end{array}$

$\text{Add [5] and [6]: }\;\underbrace{DA + BD}_{\text{This is side }c} \;=\;\frac{vy}{x} + \frac{vx}{y}\;=\;v\left(\frac{y}{x} + \frac{x}{y}\right)$

$\text{And we have: }\;c \;=\;v\cdot\frac{\overbrace{x^2+y^2}^{\text{This is }v^2}}{xy} \quad\Rightarrow\quad c \;=\;v\cdot\frac{v^2}{xy}$

. . Therefore: . $cxy \;=\;v^3$

3. Thank you very much...