2008 AMC 10B Problem 24

Three sides in a row are equal. A line that cuts the angle between two equal sides in half acts like a fold: it matches one side onto the other. So draw two such lines and look for matching triangles.

1. Draw the line that cuts angle $ABC$ in half and the line that cuts angle $BCD$ in half. Call the point where they cross $E$.
2. The halves are $70^\circ \div 2 = 35^\circ$ and $170^\circ \div 2 = 85^\circ$. So in triangle $BEC$, angle $EBC = 35^\circ$ and angle $ECB = 85^\circ$, and
$$\angle BEC = 180^\circ - 35^\circ - 85^\circ = 60^\circ.$$
3. Triangles $ABE$ and $CBE$ match. $AB = CB$, they share side $BE$, and the angles at $B$ between these sides are both $35^\circ$. Two sides and the angle between them are equal, so the triangles are congruent. Matching parts: $\angle AEB = \angle CEB = 60^\circ$ and $\angle BAE = \angle BCE = 85^\circ$.
4. Triangles $DCE$ and $BCE$ match. $DC = BC$, they share side $CE$, and the angles at $C$ are both $85^\circ$. So they are congruent, and $\angle DEC = \angle BEC = 60^\circ$.
5. Around $E$, going from $A$ to $B$ to $C$ to $D$, the three angles are $60^\circ + 60^\circ + 60^\circ = 180^\circ$. That is a straight angle, so $A$, $E$ and $D$ lie on one straight line. In other words, $E$ sits on side $AD$.
6. So angle $BAD$ is the same as angle $BAE$, which is $85^\circ$ from step 3.

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

Equal sides next to each other suggest folding along the line that halves the angle between them, which creates congruent triangles. The numbers here are set up so each angle at $E$ is $60^\circ$, and three of them make a straight line, placing $E$ right on $AD$. When a shape has several equal sides in a row, try the angle-halving lines and look for matching triangles.

Rough coordinates can confirm the answer. Put $B = (0, 0)$ and $C = (1, 0)$. Then $A$ is $1$ unit from $B$ at $70^\circ$, about $(0.34, 0.94)$. Since angle $BCD = 170^\circ$, side $CD$ leaves $C$ tilted only $10^\circ$ upward, so $D$ is about $(1.98, 0.17)$. Now $AB$ points from $A$ toward about $(-0.34, -0.94)$, and $AD$ points toward about $(1.64, -0.77)$. The angle between these directions works out to about $85^\circ$, matching (C). This takes more arithmetic than the contest time really allows, so it is best used as a check.