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
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]
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]