1981 AHSME Problems/Problem 5

Problem 5

In trapezoid $ABCD$, sides $AB$ and $CD$ are parallel, and diagonal $BD$ and side $AD$ have equal length. If $m\angle DCB=110^\circ$ and $m\angle CBD=30^\circ$, then $m\angle ADB=$

$\textbf{(A)}\ 80^\circ\qquad\textbf{(B)}\ 90^\circ\qquad\textbf{(C)}\ 100^\circ\qquad\textbf{(D)}\ 110^\circ\qquad\textbf{(E)}\ 120^\circ$


Draw the diagram using the information above. In triangle $DCB,$ note that $m\angle DCB=110^\circ$ and $m\angle CBD=30^\circ$, so $m\angle CDB=40^\circ.$

Because $AB \parallel CD,$ we have $m\angle CDB= m\angle DBA = 40^\circ.$ Triangle $DAB$ is isosceles, so $m\angle ADB = 180 - 2(40) = 100^\circ.$

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

-edited by coolmath34