given: abcd is a parallelogram. diagonals ac, bd intersect at e. prove: ae = ce and be = de. assemble the proof by dragging tiles to the statements and reasons columns. (diagram of parallelogram abcd with diagonals ac, bd intersecting at e. tiles for statements/reasons include angles: ∠cde, ∠abe, ∠ade, ∠dae, ∠bce, ∠bae, ∠bce, ∠cbe, and categories: angles, segments, triangles, statements, reasons)

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Accepted Answer

# Explanation:
## Step1: Recall properties of parallelograms
In a parallelogram $ABCD$, $AB \parallel CD$ and $AD \parallel BC$, and $AB = CD$, $AD = BC$. Also, alternate interior angles are equal when lines are parallel.
## Step2: Identify congruent triangles
To prove $AE = CE$ and $BE = DE$, we can show $\triangle ABE \cong \triangle CDE$ and $\triangle ADE \cong \triangle CBE$ using ASA or SAS.
For $\triangle ABE$ and $\triangle CDE$:
- $AB = CD$ (opposite sides of parallelogram)
- $\angle ABE=\angle CDE$ (alternate interior angles, $AB \parallel CD$, transversal $BD$)
- $\angle BAE=\angle DCE$ (alternate interior angles, $AB \parallel CD$, transversal $AC$)
By ASA, $\triangle ABE \cong \triangle CDE$, so $AE = CE$ and $BE = DE$ (corresponding parts of congruent triangles).

For the angle matching:
- $\angle ABE=\angle CDE$ (alternate interior angles)
- $\angle BAE=\angle DCE$ (alternate interior angles)
- For $\triangle ADE$ and $\triangle CBE$:
  - $AD = BC$ (opposite sides of parallelogram)
  - $\angle ADE=\angle CBE$ (alternate interior angles, $AD \parallel BC$, transversal $BD$)
  - $\angle DAE=\angle BCE$ (alternate interior angles, $AD \parallel BC$, transversal $AC$)

So the angle matches are:
$\angle ABE = \angle CDE$
$\angle BAE = \angle DCE$
$\angle ADE = \angle CBE$
$\angle DAE = \angle BCE$

# Answer:
To assemble the proof:
- For $\angle ABE$, match with $\angle CDE$ (alternate interior angles, $AB \parallel CD$, transversal $BD$)
- For $\angle BAE$, match with $\angle DCE$ (alternate interior angles, $AB \parallel CD$, transversal $AC$)
- For $\angle ADE$, match with $\angle CBE$ (alternate interior angles, $AD \parallel BC$, transversal $BD$)
- For $\angle DAE$, match with $\angle BCE$ (alternate interior angles, $AD \parallel BC$, transversal $AC$)

Then, using ASA congruence for $\triangle ABE \cong \triangle CDE$ and $\triangle ADE \cong \triangle CBE$, we get $AE = CE$ and $BE = DE$ (CPCTC - Corresponding Parts of Congruent Triangles are Congruent).

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Question

Explanation:

Step1: Recall properties of parallelograms

Step2: Identify congruent triangles

Answer: