Angle Chasing and Trigo , Cosine Law

From the OMC RMO livesolve. solved with rawlat_vanak, AdhityaMV, zvfrozel, and mathscrazy. We choose variables $\alpha = \angle BCD$, $\beta = \angle BDC$, and let the circle have unit radius. We note that $\angle AOL = \frac{1}{2} \angle AOC = \alpha$. Now \[OB \cdot (AB+CD) = (\sin \alpha + \sin (\alpha + 2\beta)) = 2 \cos \alpha \sin(\alpha + \beta) = \cos \alpha (\sin(\alpha + \beta) + \sin(\alpha + \beta)) = OL \cdot (AC + BD).\]In boards spirit, "HENCE PROVED.".

Solution: Let $\angle ABC = \alpha$ and $\angle ACB = \beta$. We know that $AC = BD$, since its an isosceles trapezium. We now need to show that \[OB(AB + CD) = OC(2\cdot AC).\][asy][asy] import olympiad; import geometry; size(180); pair A = (-2,3); pair B = (2,3); pair D = (-4,-5); pair C = (4,-5); pair O = circumcenter(A,B,C); pair L = (A+D)/2; draw(A--B--C--D--A, red); draw(A--C, orange); draw(B--D, orange); draw(O--L, orange); markscalefactor = 0.05; draw(rightanglemark(O,L,A), deepgreen); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$O$", O, dir(270)); dot("$L$", L, dir(L)); [/asy][/asy] By Sine Law on $\triangle ABC$, we get \[AB = \frac{AC}{\sin \alpha} \sin \beta. \tag{1}\]Since $\angle ACB = \angle ADB$, by a simple angle chase, we can find that $\angle CAD = 2\alpha + \beta - \pi$ and $\angle ADC = \pi - \alpha$. Apply Sine Law on $\triangle ACD$ to get \[CD = \frac{-AC}{\sin \alpha}\sin(2\alpha + \beta). \tag{2}\]Adding both the equations, we get that \begin{align*} AB + CD &= \frac{AC}{\sin \alpha} \left( \sin\beta - \sin(2\alpha + \beta) \right) \\ &= -2AC\cos(\alpha + \beta) \tag{$\bigstar$} \end{align*}where used sum to product formula. Now, finally note that $\triangle AOD$, \begin{align*} \angle AOD = 2\angle ACD \implies \angle LOD = \angle ACD = \pi - (\alpha + \beta). \end{align*}So, finally checking a simple ratio in $\triangle AOD$, we would get \[\frac{OL}{LD} = \sin(\alpha + \beta - \frac{\pi}{2}) = -\cos(\alpha + \beta).\]Write $LD = \frac{AC}{2}$ and substitute it into $(\bigstar)$. This finishes the proof. $\blacksquare$

We can use Ptolemy and cosine rule to prove it

Here's a solution without trigonometry. This is pretty clean honestly speaking, quite happy that I found it. Solution: Label $\angle LOA = \angle ACD = \alpha$. Let $M$ be the midpoint of $\overline{BC}$ and $X$ be the foot of perpendicular from $A$ onto $DC$. Now note that [asy][asy] import olympiad; import geometry; size(200); pair A = (-2,3); pair B = (2,3); pair D = (-4,-5); pair C = (4,-5); pair O = circumcenter(A,B,C); pair L = (A+D)/2; pair M = (B+C)/2; pair X = foot(A,D,C); draw(A--B--C--D--A, red); draw(A--X, orange); draw(O--A, orange); draw(L--X, orange); draw(A--C, orange); draw(L--M, orange); draw(B--D, orange); draw(O--L, orange); markscalefactor = 0.05; draw(rightanglemark(O,L,A), deepgreen); draw(rightanglemark(A,X,C), deepgreen); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$O$", O, dir(270)); dot("$L$", L, dir(L)); dot("$M$", M, dir(M)); dot("$X$", X, dir(X)); [/asy][/asy] \begin{align*} OB\cdot(AB+CD) &= OL\cdot(AC + BD) \\ \iff OB \cdot \frac{AB+CD}{2} &= OL \cdot \frac{2\cdot AC}{2} \\ \iff OA \cdot LM &= OL \cdot AC \\ \iff \cos(\theta) = \frac{OL}{OA} &= \frac{LM}{AC}. \end{align*}Therefore it suffices to show that $\cos(\theta) = \frac{LM}{AC}$. Here's the main part, look at $LMXC$. Since $L$ is the circumcenter of $\triangle AXD$, we have $MC = AL = LX$. Also, note that $\angle MCX = \angle LDX = \angle LXD$ which means $MC \parallel LX$. This means $LXMC$ is a parallelogram with $LM = XC$. Now finally, see that \[\cos(\theta) = \frac{XC}{AC} = \frac{LM}{AC}.\]This is exactly what had to be shown. $\blacksquare$ Remark: Even though you see a $\cos(\theta)$, but you can omit it altogether if you just write $\triangle AOL \sim \triangle CAX$. No trigonometry! Yay!

It is easy to see that $ABCD$ is an isosceles trapezium with $AD=BC$ and $AC=BD$. Calculate everything in terms $\angle ABD=x$ and $\angle OBC=y$. \begin{align*} OL&=R\cos(x)\\ AB&=2R\sin(2x+y)\\ AC&=2R\sin(x+y)=BD\\ CD&=2R\sin(y) \end{align*}Plugging in everything into the main equation we get \begin{align*} LHS&=4R^2\sin(x+y)\cos(x)\\ RHS&=4R^2\sin(x+y)\cos(x) \end{align*}Thus LHS=RHS and we are done.