In trapezium ABCD, AD// BC. the bisectors of angle A and angle D are d

1.	In trapezium ABCD, AD// BC. the bisectors of angle A and angle D are drawn such that they meet at a point p. find the angle ABC and angle BCD
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]

In trapezium ABCD, AD// BC. the bisectors of angle A and angle D are drawn such that they meet at a point p. find the angle ABC and angle BCD

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]

⇒

[∵ AA]

⇒

But, △AED∼△BEC [∵ Data]

⇒

[∵ AA]

⇒

[∵ Axiom 1]

=EB

EA=EB

⇒

∴AD=BC [henceproved]