Question

In the figure, ABCD is a trapezium. E and F are mid points of diagonals AC and BD respectively. Then, \(\frac{EF}{AB - DC}=\)

s\ =N

• A. True
• B. False
• C. 2
• D. 0.3

Answer 1

The correct option is B

False

Construction: Join C and F and extend CF to G, (point where it meets AB)

In $\Delta DFC$ and $\Delta BFG$,

DF = FB [F is mid-point]

$\angle DFC=\angle BFG$ [Vertically opposite angles]

$\angle CDF=\angle GBF$ [Alternate angles]

So, by A.S.A. congruence condition, $\Delta DFC\cong \Delta BFG$

$\Rightarrow$ CF = FG [C.P.C.T.C]

$\Rightarrow$ F is the mid point of CG.

In $\Delta AGC$, using mid point theorem,

$EF\parallel AG$ and $EF=\frac{1}{2}AG$

or $EF=\frac{1}{2}(AB-BG)$

But BG = CD [C.P.C.T.C]

$\Rightarrow EF=\frac{1}{2}(AB-CD)\Rightarrow \frac{EF}{AB-DC}=\frac{1}{2}=0.5$