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Here is a picture of the question:

geometry question

  • $ABCD$ is a quadrilateral.
  • $[AC]\cap[BD]=E$.
  • $|AB|=11$.
  • $|BC|=16$.
  • $|CD|=13$.
  • $|AD|=12$.
  • $|AC|=15$.
  • What is $\frac{|DE|}{|BE|}$?

This seems to be a 'plug in to formula and find' type of question; but i can't find any formulas to this situation. I know that four sides and one diagonal uniquely determine a quadrilateral, so the question is well-posed. This is supposed to be an elementary question, so i can't be sure if i'm missing something obvious or not.

[The answer is $\frac{2\sqrt2}{3}$.]

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$[ABC] = p$, a known quantity by Heron’s formula.

$[DAC] = q$, also known

$[DAE] = \dfrac {x}{15}[DAC]$; where $AE = x$

$[ABE] = \dfrac {x}{15}[ABC]$

$\dfrac {DE}{BE} = \dfrac {[DAE]}{[ABE]}$

$= \dfrac {[DAC]}{[ABC]} = \dfrac {q}{p}$

A formula is then found.

Let $AC$ be the “dividing diagonal” that divides the quadrilateral $ABCD$ into two triangles (namely $⊿BAC$ and $⊿DAC$). $BD$, the other diagonal, is then called the “intersecting diagonal”. $BD$ is being cut by $AC$ at $E$ into two parts, $BE$ - the part inside $⊿ABC$ and $DE$ - the part inside $⊿DAC$. Then,

“$\dfrac {[⊿ABC]}{(the \; part \; inside ⊿ABC)} = \dfrac {[⊿DAC]}{(the \; part \; inside ⊿DAC)}$”

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First of all you can find $\gamma=\angle DAC$ and $\beta=\angle BCA$ by the cosine rule: $$ \cos\gamma={5\over 9},\quad \cos\beta={3\over 4},\quad\hbox{whence:}\quad \sin\gamma={2\sqrt{14}\over 9},\quad \sin\beta={\sqrt{7}\over 4}. $$ Then apply the sine rule to triangles $ADE$ and $BCE$ to find: $$ DE=12\sin\gamma/\sin\alpha,\quad BE=16\sin\beta/\sin\alpha,\quad \hbox{where}\quad \alpha=\angle BEC =\angle AED. $$ By dividing these two relations $\sin\alpha$ simplifies out and you get the desired result.

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  • $\begingroup$ Minor correction: Should be $\beta=\angle BCA$. $\endgroup$ – Alistair Sep 6 '15 at 18:32
  • $\begingroup$ Corrected, thank you. $\endgroup$ – Aretino Sep 6 '15 at 19:39
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There is a simpler method which gives much more (namely all possible quadrilaterals which satisfy these conditions) and can also be used to solve many quadrilateral determination problems. We can assume that $A=(0,0),B=(16,0),C=(p_1,q_1),D=(p_2,q_2)$ where the $p$‘s and $q$‘s are to be determined. They satisfy the equations $p_2^2+q_2^2=121$, $(p_2-16)^2+q_2^2=225$, $(p_1-p_2)^2+(q_1-q_2)^2=144$ and $(p_1-16)^2+q_1^2=169$. These can be solved by hand (use the first two to solve for $p_2$ and $q_2$, then substitute these in the second two and solve for the remaining variables). Of course, one can also use a system like Mathematica to solve them at the click of a button.

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