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Angle Sum Property

In the figure...

Question

In the figure, AB=CD.Prove that BE=DE and AE=CE where E is the point of intersection of AD and BC.

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Solution

In $△ E A B$ and $△ E C D$
$A B = C D$ [Given]
$∠ B = ∠ D$ [Angles in the same segment]
$∠ A = ∠ C$ [Angles in the same segment]
$∴ △ E A B ≅ △ E C D$ [ASA criteria]
$∴ B E = D E$ [CPCT]
and $A E = C E$ [CPCT]

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Similar questions

△ A B C, D

E

A B

A C

D E ∥ B C

A D = 2.4 c m, A E = 3.2 c m, D E ∥ B C

A D = 2.4 c m, A E = 3.2 c m, D E = 2 c m

B C = 5 c m

B D

C E

Assertion: If AB = CD, then BE = DE and AE = CE, where E is the point of intersection of AD and BC.

Reason: Angles in the same segment are equal.

Which of the following options is correct?

Question 21

Fill in the blanks to make the statements true.
In the given figure, points A, B, C, D and E are collinear, such that AB = BC = CD = DE. Then,

a) AD = AB + _________

b) AD = AC + _________
c) Mid-point of AE is ______
d) Mid-point of CE is ______
e) AE = ______ AB.

Question 21

Fill in the blanks to make the statements true.
In the given figure, points A, B, C, D and E are collinear, such that AB = BC = CD = DE. Then,

a) AD = AB + _________

b) AD = AC + _________
c) Mid-point of AE is ______
d) Mid-point of CE is ______
e) AE = ______ AB.

△ A B C, D

E

A B

A C

D E ∥ B C

A D = 24 c m

A E = 3.5 c m

D E = 2 c m

B C = 5 c m

B D

C E

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