Question

E and F are respectively the midpoints of non parallel sides AB and BC of a trapezium ABCD, prove that EF || AB.​

Answer 1

$\huge\underline{\bf \red{ ❥ ᴀɴsᴡᴇʀ }}$

Let ABCD be a trapezium in AB || DC. Let E and F be the midpoints of AD and BC respectively E and F are joined.

We have to show that EF || AB.

If possible,let EF be not parallel to AB then draw EG || AB, meeting BC to G.

Now, AB || EG || DC and the transversal AD cuts them at A,E,D respectively such that AE = ED.

Also, BGC is the other transversal cutting AB,EG and DC at B, G and C respectively.

∴ BG = GC (by intercept theorem).

This shows that G is the midpoint of BC.

Hence,G must coincide with F [∴ F is the midpoint of BC].

Thus,our supposition is wrong.

Hence, EF || AB.

Answer 2

ᴀɴsᴡᴇʀ

Join CE and produce it to meet BA produced at G

In △EDC and △EAG,

⇒ ∠CED=∠GEA [ Vertically opposite angles ]

⇒ ∠ECD=∠EGA [ Alternate angles ]

⇒ ED=EA [ Since, E is the midpoint of AD ]

⇒ △EDC≅△EAG [ By AAS congruence theorem ]

⇒ CD=GA and EC=EG [ By CPCT ]

In △CGB,

E is the midpoint of CG and F is the mid-point of BC.

By mid-point theorem,

∴ EF∥AB

hence ᴘʀᴏᴠᴇᴅ

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