For a quadrilateral $ABCD$ is given that $\angle CBD=2\angle ADB$, $\angle ABD=2\angle CDB$, and $AB=CB$. Prove that $AD=CD$.
Problem
Source: X International Festival of Young Mathematicians Sozopol 2019, Theme for 10-12 grade
Tags: geometry, trigonometry
Jupiter_is_BIG
24.09.2019 15:49
*cough* trig *cough*
amar_04
24.09.2019 20:02
https://artofproblemsolving.com/community/c618937h1841423_problem_111_canada_20004
george_54
25.09.2019 14:04
Why $\Delta DCE = \Delta DZA?$
Tsikaloudakis
26.09.2019 09:30
because: \[ \left. \begin{array}{l} \frac{{CE}}{{CB}} = \frac{{DE}}{{DB}} \\ \\ \frac{{ZA}}{{AB}} = \frac{{DZ}}{{DB}} \\ \end{array} \right\}\mathop {}\nolimits_{}^{} \Rightarrow \mathop {}\nolimits_{}^{} \frac{{CE}}{{ZA}} = \frac{{DE}}{{DZ}}\mathop {}\nolimits_{}^{} \Rightarrow \mathop {}\nolimits_{}^{} ZE//AC \]
zuss77
26.09.2019 12:56
Suppose that diagonal $BD$ is inside of $ABCD$.
$\angle EBD = \angle FDB, \angle DBF = \angle BDF$
) Reflect $A,C$ about $BE,BF$. This reflections coincide in point $P$ on $BD$ ($AB=BC$).
$\angle BEP = \angle BEA = \angle BDE + \angle DBE = \angle DBF+ \angle BDF = \angle BFC = \angle BFP$
).
Place $\triangle DEB$ on top of $\triangle DFB$ so that $E,F$ coincide and rays $EB,EP,ED$ coinside with rays $FB,FP,FD$. Since $EP,FP$ equally divide $BD$ statement follows.
). Hence $AD=CD$ by symmetry.
Reflect $F \mapsto F'$ about $BD$. $BEF'P$ - cyclic. $BEF'D$ is also cyclic like isoscales trapezoid. But $B,D,P$ are collinear. Hence $E,F'$ coincide and $DEBF$ - rhombus. Result follows by symmetry.
In case when diagonal $BD$ is outside of $ABCD$ it can be shown by contradiction that $A,C$ coincide. (If $A$ outside of $\angle BCD$ , $C$ cannot be inside of $\angle BAD$).
Attachments:
Tsikaloudakis
28.09.2019 10:48
clarification: usually (in Greece at least) when we say quadrilateral ABCD we mean convex quadrilateral \[ \left. \begin{array}{l} EK = KF \\ EF//AC \\ \end{array} \right\}\mathop {}\nolimits_{}^{} \Rightarrow \mathop {}\nolimits_{}^{} AM = MC\mathop {}\nolimits_{}^{} \Rightarrow \mathop {}\nolimits_{}^{} BD \bot AC \]