3 years ago

Can I have the answers to this? Thanks

Mathematics

1 answer:

Lina20 [59]3 years ago

7 0

A) 13
b) 16
c) 2
d) 13
e) 11

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125-37=88

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There is a parallelogram ABCD with diagonals AC and BD. The diagonals AC and BD intersects each other at point E. Side AB is con

grandymaker [24]

Answer:

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Step-by-step explanation:

Given

$\square ABCD$

$\[ \lvert \[ \lvert AB =\[ \lvert \[ \lvert CD$

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Required

Which theorem shows △ABE ≅ △CDE.

From the question, we understand that:

AC and BD intersects at E.

This implies that:

$\[ \lvert \[ \lvert AE = \[ \lvert \[ \lvert EC$

and

$\[ \lvert \[ \lvert BE = \[ \lvert \[ \lvert ED$

So, the congruent sides and angles of △ABE and △CDE are:

$\[ \lvert \[ \lvert AB =\[ \lvert \[ \lvert CD$ ---- S

$\angle BAC = \angle DCA$ ---- A

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<em>Hence, the theorem that compares both triangles is the SAS theorem</em>

4 0

3 years ago

Can I have the answers to this? Thanks​

Can I have the answers to this? Thanks