Question

ABCD is a trapezium with AB || DC. E and F are points on non parallel sides AD and BC respectively such that EF || AB. Show that AE by ED = BF by FC​

Answer 1

Step-by-step explanation:

Given:

• ABCD is a trapezium in which AB || DC.
• AD and BC are non parallel sides.
• E and F are points on AD & BC.
• Also EF || AB.

To Show:

• AE/ED = BF/FC

Construction:

• Join A to C such that it intersects EF at O.

Solution: In ABCD we have

• AB || DC ......i

• EF || AB......ii

From (i) and (ii) we got EF || DC or EO || DC , similarly OF || AB

In ∆ADC , by Thales theorem

• This theorem states that :- If a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then the other two sides are divided in the same ratio.

$\implies{\rm }$ AE/ED = AO/OC.........(iii)

In ∆ABC again by Thales theorem

$\implies{\rm }$ CO/OA = CF/FB

$\implies{\rm }$ AO/OC = BF/FC.........(iv)

From (iii) and (iv) we got

➮ AE/ED = BF/FC

$\large\bold{\texttt {Proved }}$

Answer 2

Answer:

✯ Given :-

• ABCD is a trapezium with AB || DC.
• E and F are the points on parallel sides AD and BC respectively such that EF || AB.

✯ To show :-

• $\dfrac{AE}{ED}$ = $\dfrac{BF}{FC}$

✯ Solution :-

⋆ Given :

➙ AB || DC and EF || AB

» So, EF || DC (Lines parallel to the same line are parallel to each other)

➣ Now, in ∆ADC,

⇒ EG || DC (As EF || DC)

So, $\dfrac{AE}{ED}$ = $\dfrac{AG}{GC}$ ....... ❶

➣ Similarly from ∆CAB,

⇒ $\dfrac{CG}{AG}$ = $\dfrac{CF}{BF}$

⇒ $\dfrac{AG}{GC}$ = $\dfrac{BF}{FC}$ ...... ❷

➣ From the equation no ❶ and ❷ we get,

➠ $\dfrac{AE}{ED}$ = $\dfrac{BF}{FC}$

$\mapsto\boxed{\bold{\large{PROVED}}}$

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