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

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$\huge\underline\bold\orange{Question}$

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

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$\huge\underline\bold\orange{Solution}$

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.

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$\sf{\underline{\underline{\pink{Additional\:Information-}}}}$

✯︎The trapezium is a quadrilateral with two parallel sides. The parallel sides of a trapezium are called bases and the non-parallel sides of a trapezium are called legs. It is also called as a trapezoid

✯︎Trapezium a quadrilateral having one pair of opposite sides parallel is called

✯︎In trapezium KLMN , we have

• KL || NM.

✯︎The line segment joining the midpoints of non parallel sides of a trapezium is called its median

✯︎Intercept theorem is an important theorem in elementary geometry about the ratios of various line segments that are created if two intersecting lines are intercepted by a pair of parallels.

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Answer 2

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

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. Ans.