In trapezium $A B C D$ , $A B | | D C$ . $E$ and $F$ are points on non-parallel sides $A D$ and $B C$ respectively such that $E F ∥ A B$ . Prove that $\frac{A E}{E D} = \frac{B F}{F C}$ .

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Solution

$A B ∥ D C$ & $E F ∥ D C$ , therefore
$A B ∥ E F ∥ D C$
Join AC which intersects EF at G. In $△ A D C$ ,
$E G ∥ D C$ [ $∵ E F$ is the extension of EG]
$\frac{A E}{E D} = \frac{A G}{G C}$ $\to (1)$ [Lines drawn parallel to one side of triangle intersects the other two sides in distinct points. Then it is divided the other two sides in same ratio]
Similarly in $△ A B C$ , $A B ∥ G F$ , Therefore
$\frac{B F}{F C} = \frac{A G}{G C}$ $\to (2)$
From (1) & (2) ,
$\frac{A E}{E D} = \frac{B F}{F C}$

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ABCD is a trapezium with $A B ∥ D C$ . E and F are points on non-parallel sides AD and BC respectively such that $E F ∥ A B$ . Show that $\frac{A E}{E D} = \frac{B F}{F C}$ .

Q. ABCD is trapezium with AB||DC. E and F are points on non-parralel sides AD and BC respectively such that EF||AB. Show that AE/ED=BF/FC.

E

and

F

are respectively the mid points of non-parallel sides

A D, B C

of a trapezium

A B C D

Prove that

E F ∥ A B

Q. Question 12
E and F are respectively the mid-points of the non-parallel sides AD and BC of a trapezium ABCD, Prove that

E F | | A B a n d E F = \frac{1}{2} (A B + C D)

E

and

F

are respectively the midpoint of non-parallel sides

A D

and

B C

of a trapezium

A B C D

, prove that

E F

| | A B a n d E F = \frac{1}{2} (A B + C D)

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In trapezium ABCD, AB||DC. E and F are points on non-parallel sides AD and BC respectively such that EF∥AB. Prove that AEED=BFFC.

In trapezium $A B C D$ , $A B | | D C$ . $E$ and $F$ are points on non-parallel sides $A D$ and $B C$ respectively such that $E F ∥ A B$ . Prove that $\frac{A E}{E D} = \frac{B F}{F C}$ .