Let $D$ be a point on the inside of triangle $ABC$ such that $AD=CD$, $\angle DAB=70^{\circ}$, $\angle DBA=30^{\circ}$ and $\angle DBC=20^{\circ}$. Find the measure of angle $\angle DCB$.
Problem
Source: Lusophon Mathematical Olympiad 2023 Problem 2
Tags: geometry
americancheeseburger4281
28.05.2023 03:43
So trivial that it can be done by three methods:
Ceva’s trigonometric theorem on :
$$\frac{\sin \angle{CAD}}{\sin 70^\circ}\cdot\frac{\sin 30^\circ}{\sin 20^\circ}\cdot\frac{\sin\angle{BCD}}{\sin \angle{DCA}}=1
\iff\sin\angle{BCD}=2\cdot\sin20^\circ\cdot\cos20^\circ=\sin40^\circ$$Since D is inside the triangle,$\angle {BCD}=40^\circ$
Synthetic solution by drawing the circumcenter $O$ of $ADB$ and noting that: $OAD$ is equilateral, $OB=OD=AD=CD$, $\angle{ODB}=20^\circ=\angle{CBD}$, $OD$ is parallel to $BC$, $OBCD$ is a isosceles trapezoid, then $\angle{OBC}=\angle{DCB}=40^\circ$
Two law of Sines on ABD and BDC:$$
\frac{\sin 70^\circ}{BD}=\frac{\sin 30^\circ)}{AD}$$and $$\frac{sin 20}{CD}=\frac{\sin \angle{DCB}}{BD} \iff. \frac{\sin 70^\circ\cdot\sin 20^\circ}{\sin 30^\circ}=\sin \angle {DCB} =2\cdot\sin 20^\circ\cdot\cos 20^\circ=\sin 40^\circ$$
iMqther_
21.05.2024 22:32
Let $D'$ be the reflexion of D across $AB$ , with this we have $D'A=AD=AC$ and $D'B=BD$ but $\angle D'BA=\angle DBA=30^\circ$ so the $\triangle D'DB$ is equilateral $\implies$ $\angle BDD'=\angle D'DB=60^\circ$ $\implies$ $\angle AD'D=\angle D'AD=20^\circ$. Notice that $D, D'$ and $C$ are collinear and having the fact that $\triangle ADC $ is isosceles and $\angle ADD'$ is the external angle of that triangle we have $\angle ACD=\angle DAC=\frac{\angle ADD'}{2}=10^\circ$. Now by the sum of angles in $\triangle ABC$ we have $\angle ABC+\angle BAC+\angle ACB$ $\implies$ $50^\circ$ $+80^\circ$$+$ $\angle DCB$ $+10^\circ$$\implies$ $\angle DCB=180^\circ - 140^\circ=40^\circ$, and we're done. $\square$
Genius132
29.07.2024 18:29
law of sines in BCD and ABD is enough