As shown in the figure, point O is a point outside △ ABC. Take a ', B', C 'on the ray OA, ob, OC respectively, so that oa'oa = ob'ob = oc'oc = 3. Connect a'B', b'c ', c'a', and whether the obtained △ a'b'c 'is similar to △ ABC? Prove your conclusion

As shown in the figure, point O is a point outside △ ABC. Take a ', B', C 'on the ray OA, ob, OC respectively, so that oa'oa = ob'ob = oc'oc = 3. Connect a'B', b'c ', c'a', and whether the obtained △ a'b'c 'is similar to △ ABC? Prove your conclusion

It is proved by △ a ′ B ′ C ′∽ ABC. (2 points) that: from the known OA ′ OA = OC ′ OC = 3, ∠ AOC = ∠ a ′ OC ′∽ AOC ∽ a ′ OC ′, (4 points) ‖ a ′ C ′ AC = OA ′ OA = 3, similarly, B ′ C ′ BC = 3, a ′ B ′ AB = 3. (6 points) ‖ a ′ C ′ AC = B ′ C ′ BC = a ′ B ′ ab. (7 points) ‖ △ a ′ B ′ C ′∽ ABC. (8 points)