Question

4.) use what you know about congruent triangles to solve this problem. for the picture of the kite below, how would you prove that triangle bce is congruent to triangle dce if <1 is congruent to <2 and <3 is congruent to <4? (write a proof plan.)

Explanation:

Step1: Identify Common Side

Triangles \( BCE \) and \( DCE \) share side \( CE \), so \( CE = CE \) (Reflexive Property).

Step2: List Given Angles

We know \( \angle 1 \cong \angle 2 \) and need to relate angles or sides. Wait, also, in a kite, \( BC = DC \) (kite has two pairs of adjacent congruent sides). Wait, but the given is \( \angle 1 \cong \angle 2 \), \( \angle 3 \cong \angle 4 \), but for \( \triangle BCE \) and \( \triangle DCE \), let's check:
Wait, \( \angle 1 \cong \angle 2 \), \( CE \) is common, and we need another side or angle. Wait, maybe \( BC = DC \)? Wait, no, let's re - examine. The two triangles \( \triangle BCE \) and \( \triangle DCE \):

• \( \angle 1 \cong \angle 2 \) (Given)
• \( CE = CE \) (Reflexive Property)
• We need to show \( BC = DC \)? Wait, no, maybe \( BE = DE \)? Wait, no, let's think again. Wait, the problem is to prove \( \triangle BCE \cong \triangle DCE \). Let's use ASA or SAS.

Wait, \( \angle 1 \cong \angle 2 \), \( CE = CE \), and what about \( \angle BEC \) and \( \angle DEC \)? Wait, no, maybe \( BC = DC \). Wait, in a kite, adjacent sides are equal. So \( BC = DC \) (since it's a kite, \( BC \) and \( DC \) are adjacent sides from \( C \)). Then, with \( \angle 1 \cong \angle 2 \), \( CE = CE \), by SAS (Side - Angle - Side) congruence criterion, \( \triangle BCE \cong \triangle DCE \).
Wait, let's structure the proof plan:

1. Identify that \( CE \) is a common side to both \( \triangle BCE \) and \( \triangle DCE \), so \( CE\cong CE \) (Reflexive Property of Congruence).
2. Recall (or note from the kite) that \( BC\cong DC \) (a kite has two pairs of adjacent congruent sides, and \( BC \) and \( DC \) are one pair of adjacent sides from vertex \( C \)).
3. We are given that \( \angle 1\cong\angle 2 \).
4. Now, in \( \triangle BCE \) and \( \triangle DCE \):

• \( BC\cong DC \) (from kite property)
• \( \angle 1\cong\angle 2 \) (Given)
• \( CE\cong CE \) (Reflexive Property)

1. By the Side - Angle - Side (SAS) Congruence Postulate, \( \triangle BCE\cong\triangle DCE \).

Answer:

To prove that \( \triangle BCE\cong\triangle DCE \):

1. Identify the common side: Triangles \( \triangle BCE \) and \( \triangle DCE \) share the side \( CE \). By the Reflexive Property of Congruence, \( CE\cong CE \).
2. Identify the congruent side from the kite property: In a kite, adjacent sides are congruent. So, \( BC\cong DC \) (since \( BC \) and \( DC \) are adjacent sides of the kite meeting at vertex \( C \)).
3. Use the given congruent angle: We are given that \( \angle 1\cong\angle 2 \).
4. Apply the SAS Congruence Postulate: In \( \triangle BCE \) and \( \triangle DCE \), we have \( BC\cong DC \), \( \angle 1\cong\angle 2 \), and \( CE\cong CE \). By the Side - Angle - Side (SAS) Congruence Postulate, \( \triangle BCE\cong\triangle DCE \).