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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.

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 class=

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Answer:

the gram

Step-by-step explanation:

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Following are the calculation to the given points:

When the ABCD is parallelogram:

The properties of parallelogram:

[tex]\to \angle CBD = \angle ADB \\\\[/tex]

[tex]\to \angle BCA = \angle DAC \\\\[/tex]

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

[tex]\to \Delta BEC \cong \Delta AED \ \ \ \ \ \ \ \{ASA\} \\\\\to \overline {AE} \cong \overline{CE}\\\\ \to \overline{BE} \cong \overline{ED} \\\\[/tex]

When the properties of congruent triangle.

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