Question

If E and F are the mid points of non parallel sides Ad&bc respectively of trapezium Abcd then proove that EF=½(AB+CD)​

Answer 1

Appropriate Question :-

If E and F are the mid points of non - parallel sides AD & BC respectively of trapezium ABCD, then prove that EF = ½(AB+CD)

$\large\underline{\sf{Solution-}}$

Given that,

• ABCD is a trapezium with AB || CD

• E is the midpoint of AD, it means AE = ED

• F is the midpoint of BC, it means BF = FC

Construction :- Join BE, produced to G intersecting CD when produced at G.

Now, Consider triangles AEB and DEG

$\rm \: \angle \: DEG \: = \: \angle \: AEB \: \: \: \{vertically \: opposite \: angles \}$

$\rm \: DE \: = \: AE \: \: \: \{given \}$

$\rm \: \angle \: EDG\: = \: \angle \: BAE \: \: \{alternate \: interior \: angles \}$

$\rm\implies \:\rm \: \triangle \: DEG \: = \: \triangle \: AEB \: \: \: \{ASA \: congruency \}$

$\rm\implies \:EB = EG \: \: \: \: \{CPCT \}$

and

$\rm\implies \:AB = DG \: \: \: \: \{CPCT \}$

Now, in triangle BGC

F is the midpoint of BC [ given ]

EB = EG [ Proved above ]

It means, E is the midpoint of BG.

We know,

Midpoint Theorem :- This theorem states that line segment joining the midpoints of the two sides of a triangle is equals to half of the third side.

$\rm\implies \:EF \: = \: \dfrac{1}{2} \: GC$

$\rm\implies \:EF \: = \: \dfrac{1}{2} \: (GD \: + \: DC)$

$\rm\implies \:EF \: = \: \dfrac{1}{2} \: (AB \: + \: DC)$

$\rm \: [ \: \because \: AB \: = \: DG \: ]$

Hence,

$\rm\implies \boxed{\sf{  \:\rm \: \:EF \: = \: \dfrac{1}{2} \: (AB \: + \: DC) \: \: }}$

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Additional Information :-

Converse of Midpoint Theorem :-

This theorem states that if a line is drawn through the midpoint of one side of a triangle and parallel to the other side, it bisects the third side.

Answer 2

Step-by-step explanation:

ABCD is a trapezium in which $\tt A B \| C D$.

Also E and F are respectively the mid-points of sides A D and B C .

Let us join B E and produce it to meet C D produced at G, also join B D which intersects E F at O.

Now,

$\tt\triangle G E D \cong \triangle B E A \Rightarrow G E=B E \qquad\qquad [\text{By C.P.C.T}]$

Hence, E is also the mid point of G B .

Now, in $\Delta G C B$ ,E and F are respectively the mid-points of B G and B C .

$\text{\( \tt \Rightarrow \quad E F \| G C \) \: \: \: \: \: \: \: \: \: \: \: \: [By mid-point theorem]}$

But

$\text{\( \tt G C \| A B \) or \( \tt C D \| A B \) \: \: \: \: \: \: \: [Given]}$

$\tt \Rightarrow \quad E F \| A B$

In $\tt\triangle A D B, A B \| E O$ and E is the mid-point of A D.

$\text{\( \Rightarrow O \) is mid-point of \( \tt B D \) and \( \tt E O=\dfrac{1}{2} A B \) \( \ldots \) (1) [By converse of mid-point theorem]}$

$\text{In \( \tt \triangle B D C, O F \| C D \) and \( O \) is the mid-point of \( B D \).}$

$\text{say \( \tt O F=\dfrac{1}{2} C D \) \( \ldots \) (2) [By converse of mid-point theorem]}$

On adding eqn (1) & (2) we get

$\begin{aligned} \tt E O+O F & \tt=\frac{1}{2} A B+\frac{1}{2} C D \\ \\ \tt \Rightarrow \quad E F & \tt=\frac{1}{2}(A B+C D) \end{aligned}$