Question

Given that mAC=mBD. Prove CE≅EB Proof: In the same or congruent circles, congruent arcs have congruent Therefore AC= ∠C=∠, since they both intercept arc . Also, ∠AEC=∠, since those are angles. Thus by the principle of congruent triangles, CE=BE, as desired.

Answer 1

To prove CE is congruent to EB, the congruence of arcs AC and BD leads to the congruence of inscribed angles C and D. Similarly, angles AEC and BEC are congruent. By the SAS postulate, triangles AEC and BEC are congruent, confirming that CE is congruent to EB.

Step-by-step explanation:

To prove that CE≅EB, we use facts about congruent arcs and congruent triangles within circles. Firstly, it's given that mAC=mBD, which indicates that arcs AC and BD are congruent. The congruence of these arcs implies that the inscribed angles that intercept them are also congruent. Hence, ∠C and ∠D are congruent because they both intercept arc AB.

Furthermore, if we consider ∠AEC and ∠BEC as angles inscribed in a circle that intercept the same arc CE, which is a property of inscribed angles in circles, it can be stated that these angles are congruent. In the case of the triangles △AEC and △BEC, we can see that AC=BC because they are radii of the same circle, ∠AEC≅∠BEC as argued before, and CE is shared by both triangles.

By the Side-Angle-Side (SAS) postulate, if two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then the triangles are congruent. Therefore, △AEC≅△BEC implies that CE≅EB. Thus, by properties of congruent triangles, Line CE is congruent to line EB, completing the proof.