2008 AMC 10B Problem 24

Let the bisector of $\angle ABC$ and the bisector of $\angle BCD$ meet at $E$ inside the quadrilateral. Then $\angle EBC = 35^\circ$ and $\angle ECB = 85^\circ$, so in triangle $BEC$,
$$\angle BEC = 180^\circ - 35^\circ - 85^\circ = 60^\circ.$$

Because $AB = CB$ and $BE$ bisects $\angle ABC$, triangles $ABE$ and $CBE$ are congruent (SAS). Hence $\angle AEB = \angle CEB = 60^\circ$ and $\angle BAE = \angle BCE = 85^\circ$.

Similarly, $CB = CD$ and $CE$ bisects $\angle BCD$, so triangles $BCE$ and $DCE$ are congruent, giving $\angle CED = \angle CEB = 60^\circ$.

The three angles at $E$ satisfy $\angle AEB + \angle BEC + \angle CED = 180^\circ$, so $A$, $E$, $D$ are collinear: $E$ lies on side $AD$. Therefore
$$\angle BAD = \angle BAE = 85^\circ.$$

The answer is $\boxed{\textbf{(C)}\ 85}$.

Three equal consecutive sides invite folding: reflecting over the bisector of the angle between two equal sides swaps them, and the intersection $E$ of two such bisectors is equidistant from all four vertices' "fold images." The specific numbers make $\angle BEC = 60^\circ$, so three copies fill a straight angle and $E$ falls on $AD$. When a quadrilateral has equal adjacent sides, draw the bisectors of the angles between them and look for congruent triangles.

Coordinates with $B = (0,0)$, $C = (1,0)$: $A = (\cos 70^\circ, \sin 70^\circ)$ and $D = (1 + \cos 10^\circ, \sin 10^\circ)$. The dot product of $\vec{AB}$ and $\vec{AD}$ gives $\cos\angle BAD \approx 0.087$, so $\angle BAD \approx 85^\circ$; a reasonable calculator-free estimate still singles out (C).

• Trying to angle-chase through triangle ACD without a construction, which leaves the split of the 120 degrees between angles A and D undetermined.
• Concluding $\angle BAD = 55^\circ + 5^\circ = 60^\circ$ or similar by adding base angles of the two isosceles triangles that do not share the vertex $A$.