Final answer:
By recognizing that AOC and BOD are diameters of a circle with O as the center, we can conclude that triangles ABD and DCA have right angles at O and share the common side AD. AO = CO and BO = DO as they are radii of the circle, thus by the RHA theorem, triangle ABD is congruent to triangle DCA.
Step-by-step explanation:
The question asks us to prove that triangles ABD and DCA are congruent using the Right angle-Hypotenuse-Angle (RHA) congruence theorem. The information provided suggests that AOC and BOD are diameters of a circle with the center at O, which implies that all angles at the point O, formed by these diameters and any other point on the circumference, will be right angles. Therefore, both triangles ABD and DCA have a right angle at O.
Next, since AOC and BOD are diameters, the segments AO, CO, BO, and DO are all radii of the circle and therefore equal in length. This means AO = CO and BO = DO, which establishes that triangles ABD and DCA each have two sides that are equal in length; AO = CO (or DO) and BO = DO (or AO) for triangles ABD and DCA, respectively.
Lastly, the shared side AD is a common side for both triangles. Now, we have two triangles with two sides equal and the included angle equal (the right angle at O). Hence, by RHA (Right angle-Hypotenuse-Angle) congruence theorem, triangle ABD is congruent to triangle DCA.