∠aeb ≅ ∠cdb
∠bed ≅ ∠bde
which of the following relationships proves why △aed and △cde are congruent?
sss
aas
sas
asa

Question

Accepted Answer

# Brief Explanations:
1. First, analyze the given information: We know $ \angle AEB \cong \angle CDB $ and $ \angle BED \cong \angle BDE $. From $ \angle BED \cong \angle BDE $, triangle $ BED $ is isosceles, so $ BE = BD $. Also, $ ED $ is a common side to both $ \triangle AED $ and $ \triangle CDE $.
2. Now, let's check the angles: $ \angle AED=\angle AEB + \angle BED $ and $ \angle CDE=\angle CDB+\angle BDE $. Since $ \angle AEB \cong \angle CDB $ and $ \angle BED \cong \angle BDE $, we have $ \angle AED\cong\angle CDE $.
3. We also know $ ED = DE $ (common side) and from the isosceles triangle $ BED $, we can infer about the sides or use angle - side - angle. Wait, let's re - evaluate. Wait, actually, let's look at the triangles $ \triangle AED $ and $ \triangle CDE $:
   - $ ED $ is common, so $ ED = DE $.
   - We have $ \angle AED\cong\angle CDE $ (as shown above).
   - Also, from $ \angle BED\cong\angle BDE $, we know that $ \triangle BED $ is isosceles, but more importantly, let's check the other angle. Wait, maybe a better approach: Let's consider the ASA (Angle - Side - Angle) criterion. The ASA criterion states that if two angles and the included side of one triangle are congruent to two angles and the included side of another triangle, the triangles are congruent.
   - We have $ \angle AED\cong\angle CDE $, $ ED = DE $ (common side), and we can find another pair of angles. Wait, actually, let's correct: Wait, the triangles $ \triangle AED $ and $ \triangle CDE $:
     - $ \angle ADE $ and $ \angle CED $: Wait, no. Wait, let's start over. We know that $ \angle BED\cong\angle BDE $, so $ BE = BD $. Also, $ \angle AEB\cong\angle CDB $, $ \angle ABE\cong\angle CBD $ (vertical angles). So $ \triangle ABE\cong\triangle CBD $ by AAS, which gives $ AE = CD $. Now, in $ \triangle AED $ and $ \triangle CDE $:
       - $ AE = CD $ (from $ \triangle ABE\cong\triangle CBD $).
       - $ \angle AED=\angle CDE $ (as $ \angle AEB+\angle BED=\angle CDB + \angle BDE $).
       - $ ED = DE $ (common side).
     - So by SAS (Side - Angle - Side), we have two sides and the included angle? Wait, no. Wait, $ AE = CD $, $ ED = DE $, and $ \angle AED=\angle CDE $. So the sides $ AE $ and $ ED $ with included angle $ \angle AED $ in $ \triangle AED $, and sides $ CD $ and $ DE $ with included angle $ \angle CDE $ in $ \triangle CDE $. So that's SAS. Wait, but let's check the options again. Wait, maybe I made a mistake. Wait, the other approach: The ASA (Angle - Side - Angle) criterion: If we have two angles and the included side. Let's see, $ \angle ADE $ and $ \angle CED $: No, maybe the correct way is:
       - We know that $ \angle BED\cong\angle BDE $, so $ BE = BD $. Also, $ \angle AEB\cong\angle CDB $, $ \angle ABE\cong\angle CBD $ (vertical angles), so $ \triangle ABE\cong\triangle CBD $ (AAS), so $ AE = CD $. Now, in $ \triangle AED $ and $ \triangle CDE $:
         - $ \angle AED=\angle AEB+\angle BED $, $ \angle CDE=\angle CDB + \angle BDE $. Since $ \angle AEB\cong\angle CDB $ and $ \angle BED\cong\angle BDE $, $ \angle AED\cong\angle CDE $.
         - $ ED = DE $ (common side).
         - $ AE = CD $ (from $ \triangle ABE\cong\triangle CBD $).
       - Wait, no, that's S - A - S? Wait, $ AE = CD $, $ \angle AED=\angle CDE $, $ ED = DE $. So the sides $ AE $ and $ ED $ with included angle $ \angle AED $ in $ \triangle AED $, and sides $ CD $ and $ DE $ with included angle $ \angle CDE $ in $ \triangle CDE $. So that's SAS. But wait, maybe the correct answer is ASA? No, let's think again. Wait, the key is:
         - $ ED $ is common.
         - $ \angle AED\cong\angle CDE $ (as shown).
         - $ \angle ADE\cong\angle CED $? No, wait, from $ \angle BED\cong\angle BDE $, we know that $ \triangle BED $ is isosceles, so $ BE = BD $, and from $ \triangle ABE\cong\triangle CBD $, we have $ AE = CD $. Then, in $ \triangle AED $ and $ \triangle CDE $:
           - $ \angle AED\cong\angle CDE $ (angle), $ ED = DE $ (side), $ \angle ADE\cong\angle CED $? No, maybe the correct criterion is ASA. Wait, no, let's check the definitions:
             - ASA: Two angles and the included side.
             - SAS: Two sides and the included angle.
             - AAS: Two angles and a non - included side.
             - SSS: Three sides.
           - We have $ \angle AED\cong\angle CDE $ (angle), $ ED = DE $ (side), and we can find that $ \angle ADE\cong\angle CED $? Wait, no, let's look at the given angles again. The problem gives $ \angle AEB\cong\angle CDB $ and $ \angle BED\cong\angle BDE $. So $ \angle AED=\angle AEB+\angle BED $, $ \angle CDE=\angle CDB+\angle BDE $, so $ \angle AED\cong\angle CDE $. Also, $ ED $ is common, and we can show that $ \angle ADE\cong\angle CED $? No, maybe the correct answer is ASA. Wait, no, the correct answer is ASA? Wait, no, let's take a step back. The triangles $ \triangle AED $ and $ \triangle CDE $:
             - $ \angle AED\cong\angle CDE $ (angle), $ ED = DE $ (side), $ \angle ADE\cong\angle CED $? No, maybe the answer is ASA. Wait, I think I made a mistake earlier. Let's re - examine the diagram. Points $ E $ and $ D $ are on the base, $ B $ is the intersection of $ AC $ and $ BD $ (wait, no, the diagram shows $ A $ connected to $ E $ and $ B $, $ C $ connected to $ D $ and $ B $, $ E $ and $ D $ are connected, and $ A $ and $ C $ are connected? Wait, no, the diagram is a quadrilateral $ AEDC $ with $ E $ and $ D $ at the bottom, $ A $ and $ C $ at the top, and $ B $ is the intersection of $ AD $ and $ CE $. So $ ED $ is the base. So $ \triangle AED $ and $ \triangle CDE $ share the side $ ED $. We know $ \angle AED\cong\angle CDE $ (from the given angles $ \angle AEB\cong\angle CDB $ and $ \angle BED\cong\angle BDE $), $ ED = DE $ (common side), and we can find that $ \angle ADE\cong\angle CED $? No, wait, the correct criterion is ASA. Wait, no, the answer is ASA? Wait, no, let's check the options again. The correct answer is ASA? No, I think the correct answer is ASA. Wait, no, let's think about the angles:
               - $ \angle AED\cong\angle CDE $ (angle), $ ED = DE $ (side), $ \angle ADE\cong\angle CED $ (angle). So two angles and the included side (ED), so ASA. Wait, but earlier I thought SAS, but maybe I was wrong. Wait, the key is that $ \angle AED$ and $ \angle CDE $ are angles, $ ED $ is the included side, and then $ \angle ADE $ and $ \angle CED $ are the other angles. So the correct congruence criterion is ASA. Wait, no, let's check the given information again. The problem is asking which relationship proves $ \triangle AED\cong\triangle CDE $. We have:
                 - $ \angle AED\cong\angle CDE $ (as $ \angle AEB+\angle BED=\angle CDB+\angle BDE $ and $ \angle AEB\cong\angle CDB $, $ \angle BED\cong\angle BDE $).
                 - $ ED = DE $ (common side).
                 - $ \angle ADE\cong\angle CED $? Wait, no, from $ \angle BED\cong\angle BDE $, we know that $ BE = BD $, and from $ \triangle ABE\cong\triangle CBD $ (AAS, because $ \angle AEB\cong\angle CDB $, $ \angle ABE\cong\angle CBD $, $ BE = BD $), we get $ AE = CD $. Then in $ \triangle AED $ and $ \triangle CDE $: $ AE = CD $, $ \angle AED=\angle CDE $, $ ED = DE $. So that's SAS (side - angle - side). But the options include SAS. Wait, maybe I made a mistake in the angle calculation. Let's re - do the angle calculation:
                   - $ \angle AED=\angle AEB+\angle BED $
                   - $ \angle CDE=\angle CDB+\angle BDE $
                   - Given $ \angle AEB\cong\angle CDB $ and $ \angle BED\cong\angle BDE $, so $ \angle AED\cong\angle CDE $ (by angle addition postulate).
                   - $ ED $ is common, so $ ED = DE $.
                   - We also have $ AE = CD $ (from $ \triangle ABE\cong\triangle CBD $ by AAS, since $ \angle AEB\cong\angle CDB $, $ \angle ABE\cong\angle CBD $ (vertical angles), and $ BE = BD $ (from $ \triangle BED $ being isosceles as $ \angle BED\cong\angle BDE $)).
                   - So in $ \triangle AED $ and $ \triangle CDE $: $ AE = CD $, $ \angle AED=\angle CDE $, $ ED = DE $. So this is SAS (side - angle - side), because we have two sides ($ AE = CD $, $ ED = DE $) and the included angle ($ \angle AED=\angle CDE $) congruent. But wait, the answer options have SAS. But earlier I was confused. Wait, no, let's check the ASA again. If we consider $ \angle AED\cong\angle CDE $, $ ED = DE $, and $ \angle ADE\cong\angle CED $, then it's ASA. But do we know $ \angle ADE\cong\angle CED $? From $ \triangle BED $ being isosceles, $ BE = BD $, and $ \triangle ABE\cong\triangle CBD $, we have $ AE = CD $, and $ ED $ is common. So by SSS, would that work? No, because we only know two sides? Wait, no, $ AE = CD $, $ ED = DE $, and what about $ AD $ and $ CE $? No, we don't know that. So the correct criterion is SAS, because we have two sides ($ AE = CD $, $ ED = DE $) and the included angle ($ \angle AED=\angle CDE $) congruent. Wait, but the answer is SAS? Wait, no, the correct answer is ASA? I think I made a mistake. Let's check the definitions again:
                     - SAS: If two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, the triangles are congruent.
                     - ASA: If two angles and the included side of one triangle are congruent to two angles and the included side of another triangle, the triangles are congruent.
                     - AAS: If two angles and a non - included side of one triangle are congruent to two angles and the corresponding non - included side of another triangle, the triangles are congruent.
                     - SSS: If three sides of one triangle are congruent to three sides of another triangle, the triangles are congruent.
                   - We have $ \angle AED\cong\angle CDE $ (angle), $ ED = DE $ (side), and we can show that $ \angle ADE\cong\angle CED $ (angle). How? Since $ \triangle BED $ is isosceles ($ \angle BED\cong\angle BDE $), $ BE = BD $. Also, $ \triangle ABE\cong\triangle CBD $ (AAS: $ \angle AEB\cong\angle CDB $, $ \angle ABE\cong\angle CBD $, $ BE = BD $), so $ AE = CD $. Then in $ \triangle AED $ and $ \triangle CDE $, we have $ AE = CD $, $ ED = DE $, and $ AD = CE $? No, we don't know that. Wait, I think the correct answer is ASA. Wait, no, let's look at the diagram again. The triangles $ \triangle AED $ and $ \triangle CDE $ share the side $ ED $. We have $ \angle AED\cong\angle CDE $ (from the given angle additions) and $ \angle ADE\cong\angle CED $ (because $ \triangle AED $ and $ \triangle CDE $ are related through the isosceles triangle and the congruent triangles above). And $ ED $ is the included side. So that's ASA. But I'm getting confused. Wait, the correct answer is ASA? No, the answer is SAS? Wait, the problem is from a geometry question, and the correct answer is ASA? No, let's check with the given information. The key is that $ \angle BED\cong\angle BDE $ implies $ BE = BD $, $ \angle AEB\cong\angle CDB $ and $ \angle ABE\cong\angle CBD $ (vertical angles) implies $ \triangle ABE\cong\triangle CBD $ (AAS), so $ AE = CD $. Then in $ \triangle AED $ and $ \triangle CDE $: $ AE = CD $, $ ED = DE $, $ \angle AED=\angle CDE $ (as $ \angle AEB+\angle BED=\angle CDB+\angle BDE $). So this is SAS, because the angle is between the two sides ($ AE $ and $ ED $ in $ \triangle AED $, $ CD $ and $ DE $ in $ \triangle CDE $). So the correct answer is SAS. But wait, the options have SAS. So the answer is SAS.
# Answer:
SAS (the option with "SAS" text, e.g., if the option is labeled as "SAS" among the choices, then the answer is the option with "SAS")

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