Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.5 National olympiad Prove it Japan

Suppose the points $A$ , $B$ , $C$ , $D$ are located on the circumference of a circle in this order as indicated in the figure below.
Suppose the angle formed by the line tangent to the circle at $B$ and the line $AB$ is $30^\circ$ , and that formed by the line tangent to the circle at $C$ and the line $CD$ is $10^\circ$ . Suppose further that the lines $AB$ and $DC$ are parallel and they are located on the opposite sides from each other with respect to the center of the circle. Determine the magnitude of the angle $\angle BDC$ .

Solution

Since $AB \parallel DC$ and since $\angle DCA$ and $\angle DBA$ are angles subtended by the same arc $\text{AD}$ at the points $C$ and $B$ on the circle, we have $\angle BDC = \angle DBA = \angle DCA$ .

Since the angle subtended by an arc $\text{AB}$ at the point $C$ on the circle equals the angle formed by the tangent line to the circle at $B$ and the chord $AB$ , which is $30^\circ$ , we have $\angle ACB = 30^\circ$ .

Similarly, we have $\angle DBC = 10^\circ$ .

Therefore, the sum of the inner angles of the triangle $BCD$ equals
$\angle BDC + \angle DCA + \angle ACB + \angle DBC = 2\angle BDC + 30^\circ + 10^\circ,$
from which we get
$\angle BDC = \frac{180^\circ - (10^\circ + 30^\circ)}{2} = 70^\circ.$

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