A,B,C and D are four points on a circle in that cyclic order. If $AD=BD=50 \sqrt3 cm$, $AC=106.8cm$ and $∠CAD=30$, what is the length, in cm, of BC?

Question

A,B,C and D are four points on a circle in that cyclic order. If $AD=BD=50 \sqrt3 cm$, $AC=106.8cm$ and $∠CAD=30$, what is the length, in cm, of BC?

Since ABCD is cyclic there would be two similar triangles, I tried using their equal ratios and algebra but it got me nowhere

Taken from 2014 IMC

Answer 1

Let's denote the cross of AC and DB as E. Drop a perpendicular from D to AC at F; in right triangle ADF we have:

$DF=\frac{AD}{2}=25\sqrt 3$

$AF=AD\times Cos(30)=75$

Therefore:

$FC=106.8-75=31.8$

In right triangle DFC we have:

$DC=\sqrt{(25\sqrt 3)^2+31.8^2}=53.72$

$Sin(\angle DCF)=\frac{25\sqrt3}{53.72}$

⇒$\angle DCF≈53.7^o $

Therefore:

$\angle ADC=180-(30 +53.7)≈96.3$

⇒$Arc (AC)=360-2\times 96.3=167.4$

D is the midpoint of the arc AB and C is between A and B and we have:

$arc CD=\frac{|arc AC-arc CB|}{2}=\frac{187.4-arc CB}{2}$

⇒ $arc CB=47.4$

⇒ $\angle CDE=(47.4)/2=23.7$

⇒ $\angle AED=53.7+23.7=77.4$

⇒ $FE=\frac{25\sqrt 3}{\ tan 77.4}≈9.67$

⇒ $AE=AF+FE=75+9.7=84.7$

Triangles ADE and BCE are similar and we have:

$\frac{CB}{AD}=\frac{EB}{AE}$

⇒ $CB=\frac{50\sqrt 3 \times 42.3}{84.67}=43.26$

Answer 2

Let $BC=x$.

Since $\measuredangle DAC=\measuredangle DBC=30^{\circ},$ by law of cosines we obtain: $$x^2+(50\sqrt3)^2-2x\cdot50\sqrt3\cdot\frac{\sqrt3}{2}=106.8^2+(50\sqrt3)^2-2\cdot106.8\cdot50\sqrt3\cdot\frac{\sqrt3}{2}$$ or $$x^2-150x+4613.76=0,$$ which gives $$x=43.2$$ or $$x=106.8.$$ Can you end it now?

I got that the second option is impossible.