Math, asked by muskan7692, 4 months ago

In a trapezium ABCD wherw AB is parallel to CD, E is the midpoint of BC, prove that triangle AED=1/2 trapezium ABCD​

Answers

Answered by jashankamboj1717
7

Answer:

ANSWER

In the □ABCD

AB || DC

△AED∼△BEC

To Prove: AD = BC

Proof: Compare △EDC and △EBA

∠EDC=∠EBA [∵ AB || DC and Alternate Angles]

∠ECD=∠EAB

∴△EDC∼△EBA [∵ Equiangular triangles]

EB

ED

=

EA

EC

[∵ AA]

EC

ED

=

EA

EB

But, △AED∼△BEC [∵ Data]

EC

ED

=

EB

EA

=

BC

AD

[∵ AA]

EA

EB

=

EB

EA

[∵ Axiom 1]

EA

2

=EB

2

EA=EB

EB

EA

=

BC

AD

=1

∴AD=BC [henceproved]

Answered by Hp10Sourav
1

Answer:

ANSWER

In the □ABCD

AB || DC

△AED∼△BEC

To Prove: AD = BC

Proof: Compare △EDC and △EBA

∠EDC=∠EBA [∵ AB || DC and Alternate Angles]

∠ECD=∠EAB

∴△EDC∼△EBA [∵ Equiangular triangles]

EB

EE = EA EC

[∵ AA]

EC

ED

=

EA

EB

But, △AED∼△BEC [∵ Data]

EC

ED

=

EB

EA

=

BC

AD

[∵ AA]

EA

EB

=

EB

EA

[∵ Axiom 1]

EA

2

=EB

2

EA=EB

EB

EA

=

BC

AD

=1

∴AD=BC [henceproved]

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