Geometry problem: Given a fixed acute triangle $\bigtriangleup{ABC}$ with $\overline{AB}=13$ and $\overline{BC}=7$. $D$ and $E$ are points on sides $\overline{AB}$ and $\overline{AC}$ respectively such that $\overline{BD}=\overline{BC}$ and angle $\angle DEB= \angle CEB$. Find the product of all possible values of the length of $\overline{AE}$.
The image of the triangle is not given but my teacher gave the image to us as a hint. At the end of the course he even gave us the solution but there was a point I did not understand. I got the answer and the solution. I look forward to seeing other solutions, please.
Solution: Method of false position, 2 cases provided.
Case 1: $\bigtriangleup{BED}$ $\cong$ $\bigtriangleup{BEC}$
$\angle{DBE}=\angle{DBE}$ (corr. $\angle$s, $\cong$$\bigtriangleup$s)
$AE\over{EC}$=$AB\over{BC}$=$13\over{7}$ ($\angle$ bisector thm)
AE=${13\overline{AC}}\over13+7$=${13\overline{AC}}\over20$
Case 2: $\bigtriangleup{BED}$ $\cong$ $\bigtriangleup{BEC}$ is false
B,C,E,D are concyclic. (????)
$\overline{AD}\times\overline{AB}=\overline{AE}\times\overline{AC}$ (????)
$6\times13=\overline{AE}\times\overline{AC}$
$\overline{AE}$$=$${78}\over\overline{AC}$
Product of possible cases=
${78}\over\overline{AC}$$\times$${13\overline{AC}}\over20$
=50.7 units
The parts I dont understand:
- Why is B,C,E,D concyclic if the triangles are not congruent?
- What is that property of cyclic quadrilateral?
I also hope to see new solutions pop out!
The idea is that you have two triangles that have two pairs of equal sides and a pair of equal angles ($BE$ is shared, $BC = BD, \angle CEB = \angle DEB$), but the triangles do not have to be congruent. By the sine rule, $\sin \angle BCE = \sin \angle BDE$, hence those angles are either equal or supplementary.
Case 1 is when $\angle BCE = \angle BDE$, then $BE$ is the bisector of $\angle ABC$.
Case 2 is when $\angle BCE + \angle BDE = \pi$. For a cyclic quadrilateral, the sum of the opposite angles is $\pi$, since the sum of the corresponding central angles is $2 \pi$, and an inscribed angle is one half of the central angle. The converse is also true, that is, if the opposite angles are supplementary, then the quadrilateral is cyclic. The last step is applying the intersecting secants theorem.