Finding segment in a right triangle.

Question

Here is the picture of the question:

• $ABC$ is a right triangle.
• $m(CBA)=90^\circ$.
• $m(BAD)=2m(DAC)=2\alpha$.
• $D$ is a midpoint of $[BC]$.
• $E$ is a point on $[AD]$.
• $m(BED)=90^\circ$.
• $|DE|=3$.
• What is $|AB|=x$?

Tried lots of things which gives me some trigonometric identities, but none of them led me to the solution which is $x=6$. And even if it did, i prefer more geometric methods (all solutions are welcome though). Note that this could be an easy question, and i could probably missing something obvious.

Answer 1

(1) The brown dotted line is the extension of $BE$.

(2) The green dotted line ($AC’$) is the angle bisector of $∠BAE$ such that $\beta_1 = \beta_2 = \beta$. Another obvious fact is $\beta ‘ = 2\beta$.

(3) The blue dotted line is the perpendicular bisector of $AB$ cutting $AB$ and $AC$ at $P$ and $Q$ respectively.

By intercept theorem, $AQ = QC$.

By midpoint theorem, $QD = BP = PA$.

$L$ is point on $BC$ such that $EL \bot BC$. $EL$ is extended to cut $AC’$ at $N$. It should also be clear that $\angle 1 = \angle 2 = \beta’$..

The line $AEDM$ is the axis of symmetry for $\triangle ACC’$. Then, $\angle AMC = 90^\circ$. This means M is also on the circum-circle (centered at Q) of $\triangle ABC$.

Therefore, $\triangle BED \equiv \triangle CMD \Rightarrow DE = DM$.

$QA = QM \Rightarrow \beta_3 = \beta$. $\beta_4 = \angle 2 - \beta_3 = \beta \Rightarrow QD = DM$.

Result follows.

Answer 2

This is one approach. I don't pretend it's the nicest.

Let $u = \tan\alpha$. Then

$$ \tan2\alpha = \frac{2u}{1-u^2} $$ $$ \tan3\alpha = \frac{u(3-u^2)}{1-3u^2} $$

Since $\tan3\alpha = 2\tan2\alpha$, we have

$$ \frac{4u}{1-u^2} = \frac{u(3-u^2)}{1-3u^2} $$ $$ (1-u^2)(3-u^2) = 4(1-3u^2) $$ $$ 3-4u^2+u^4 = 4-12u^2 $$ $$ 1-u^4 = 8u^2 $$

Now let $y = BE$. By similar triangles, we have

$$ \frac{3}{y} = \frac{\sqrt{9+y^2}}{x} = \tan2\alpha = \frac{2u}{1-u^2} $$

Square all terms to obtain

$$ \frac{9}{y^2} = \frac{9+y^2}{x^2} = \left(\frac{2u}{1-u^2}\right)^2 = \frac{4u^2}{1-2u^2+u^4} $$

Observe that the outer pair of the equality above gives us

$$ \frac{9+y^2}{y^2} = \frac{9}{y^2}+1 = 1+\frac{4u^2}{1-2u^2+u^4} = \frac{1+2u^2+u^4}{1-2u^2+u^4} = \left(\frac{1+u^2}{1-u^2}\right)^2 $$

Multiplying both ends by $\frac{x^2}{9+y^2} \cdot \frac{y^2}{9} = \left(\frac{1-u^2}{2u}\right)^2\left(\frac{1-u^2}{2u}\right)^2$ yields

\begin{align} \frac{x^2}{9} & = \left(\frac{1+u^2}{2u}\right)^2 \left(\frac{1-u^2}{2u}\right)^2 \\ & = \left(\frac{1-u^4}{4u^2}\right)^2 \\ & = \left(\frac{8u^2}{4u^2}\right)^2 \qquad \longleftarrow 1-u^4 = 8u^2 \\ & = 2^2 = 4 \end{align}

So $x^2 = 36$ and $x = 6$.

Neat question. One almost feels as though the simple and straightforward answer must be obtainable with a correspondingly simple and straightforward approach. Thus far, though, I haven't seen it. Anybody?