In the figure above, angle BCD measures 90 degrees, angle ABC measures

Draw segment DB. This separates the quadrilateral into two triangles: BDC, which is 30-60-90, and ADB, which is isosceles (because AD and AB are congruent). Based on known ratios, angle DBC is 60 degrees and angle BDC 30 degrees. Because we were given ABC is 120 degrees and have reasoned that DBC is 60 degrees, we know angle ABD is also 60 degrees.

Now we return to triangle ABD. Congruent sides are opposite congruent angles. Therefore, angle ADB is 60 degrees. Angles ADB and BDC collectively form angle ADC. Add ADB and BDC. 60 + 30 = 90. The answer is B.

VERITAS PREP OFFICIAL SOLUTION:

As a general rule with geometry questions, when in doubt look for (or draw) right triangles! In this instance, when you're given the sides y and y√3 your mind should immediately start thinking of the 30-60-90 side ratio, especially when you're told that angle BCD already gives you the 90 degree angle. Draw a line connecting points B and D as shown in blue below, and you can complete that diagram for the 30-60-90 triangle you've drawn.

That also means that, since angle DBC measures 60 and the larger angle ABC measures 120, that angle ABD has to be 60 as well (also shown in blue). What does that mean for the other triangle in the new diagram, triangle ABD? Since you already know it's isosceles (AD = AB) and you know that one angle is 60, that means that all angles are 60 and it must be an equilateral triangle. Therefore angle ADB = 60 and you already know from the 30-60-90 triangle that angle BDC = 30. Add those two components of angle ADC and you have a 90-degree angle.

Answer = B = 90

\(\triangle\) ADB is an equilateral triangle & \(\triangle\)DCB is 30-60-90

So \(\angle\) ADC = 90

\(DB^{2} = BC^{2} + CD^{2}\)
\(DB^{2}= y^{2} + 3y^{2}\)
\(DB^{2} = 4y^{2}\)
\(DB = 2y\)

DB, DC, and CB are 2y, \(\sqrt{3}\)y, and y; Note that (2:\(\sqrt{3}\):1) are for triangle with \((90^{\circ}:60^{\circ}:30^{\circ})\)
So, \(BCD=90^{\circ}\); \(CBD=60^{\circ}\); \(BDC=30^{\circ}\)

Next, \(ABC=120^{\circ}\); So \(ABD=ABC-CBD=120^{\circ}-60^{\circ}=60^{\circ}\)
AB = AD so \(ADB=ABD=60^{\circ}\)

Hence, \(ADC=BDC+ADB=30^{\circ}+60^{\circ}=90^{\circ}\)

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