Problem 1669

Solution:

We have $A \widehat{C} D = C \widehat{D} O$, because they are alternate interior angles with respect to the parallels $A C$ and $D O$; moreover $A \widehat{C} D = A \widehat{B} D$, since they subtend the same arc of the circle. Triangle $D O B$ is formed by two radii, and is therefore isosceles; from this we obtain the congruence of its base angles $O \widehat{D} B$ and $O \widehat{B} D$. Hence, summing up, $C \widehat{D} O = A \widehat{C} D = A \widehat{B} D = O \widehat{D} B$: $D O$ is the bisector of $C \widehat{D} B$.

$D \widehat{C} B = D \widehat{A} B$, since they subtend the same arc. Moreover $A \widehat{O} D = 2 A \widehat{B} D$ (central angle and inscribed angle subtending the same arc), and hence $A \widehat{O} D = C \widehat{D} B$. But then triangles $A O D$ and $C D B$ are similar by the first criterion of similarity (three congruent angles).

Alternatively, once the congruence of one of the two aforementioned angles has been shown, one can proceed as follows. $D O$ is both the bisector and the altitude of triangle $C D B$ ($D O$ is parallel to $A C$ and $A C$ is perpendicular to $C B$, since $A \widehat{C} B$ subtends a diameter). Therefore triangle $C D B$ is isosceles on base $B C$; but triangle $A D O$ is also isosceles, having two radii as sides. From this it follows that the two triangles are similar by the second criterion of similarity.