Question

Given: ABCD is a parallelogram.
Diagonals AC, BD intersect at E.
Prove: AE = CE and BE = DE
B.
С
E
A
D
Assemble the proof by dragging tiles to
the Statements and Reasons columns.

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Following are the calculation to the given points:

When the ABCD is parallelogram:

The properties of parallelogram:

$\to \angle CBD = \angle ADB$

$\to \angle BCA = \angle DAC$

When the Two-lines are parallel and alternate interior angles are equal:

$\to \Delta BEC \cong \Delta AED \ \ \ \ \ \ \ \{ASA\} \\\\\to \overline {AE} \cong \overline{CE}\\\\ \to \overline{BE} \cong \overline{ED}$

When the properties of congruent triangle.

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