Question

"Question 8 XY is a line parallel to side BC of a triangle ABC. If BE || AC and CF || AB meet XY at E and E respectively, show that ar (ABE) = ar (ACF)

Class 9 - Math - Areas of Parallelograms and Triangles Page 163"

Answer 1

If a parallelogram and a triangle are on the same base and between the same parallels then area of the triangle is half the area of the parallelogram.

Parallelograms on the same base and between the same parallels are equal in area.

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Given,In ∆ABC, in which XY || BC, BE || AC and CF || AB i.e CF||XB.

To show,
ar(ΔABE) = ar(ΔAC)

Proof:
EY || BC (XY || BC) — (i)
also,
BE∥ CY (BE || AC) — (ii)

From (i) and (ii),
BEYC is a parallelogram.

(Both the pairs of opposite sides are parallel.)

Similarly,We can prove that BXFC is a parallelogram.

Parallelograms on the same base BC and between the same parallels EF and BC.

 ar(BEYC) = ar(BXFC)....(iii)

 

(Parallelograms on the same base BC and between the same parallels EF and BC)

 
Also,
△ABE and parallelogram BEYC are on the same base BE and between the same parallels BE and AC.

ar(△ABE) = 1/2ar(BEYC) — (iv)

Similarly,
△ACF & ||gm  BXFC on the same base CF and between the same parallels CF and AB.

ar(△ ACF) = ½ ar(BXFC) — (v)

From (iii), (iv) and (v),
ar(△ABE) = ar(△ACF)

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Hope this will help you...

 

Answer 2

Answer:Given,In ∆ABC, in which XY || BC, BE || AC and

CF || AB i.e CF||XB.

To show,

ar(ΔABE) = ar(ΔAC)

Proof:

EY || BC (XY || BC) — (i)

also,

BE∥ CY (BE || AC) — (ii)

From (i) and (ii),

BEYC is a parallelogram.

(Both the pairs of opposite sides are

parallel.)

Similarly,We can prove that BXFC is a parallelogram.

Parallelograms on the same base BC and between

the same parallels EF and BC.

 ar(BEYC) = ar(BXFC)....(iii)

 

(Parallelograms on the same base BC and between

the same parallels EF and BC)

 

Also,

△ABE and parallelogram BEYC are on the same base BE and between the same parallels

BE and AC.

ar(△ABE) = 1/2ar(BEYC) — (iv)

Similarly,

△ACF &

||gm  BXFC on the same base CF and

between the same parallels CF and AB.

ar(△ ACF) = ½ ar(BXFC) — (v)

From (iii), (iv) and (v),

ar(△ABE) = ar(△ACF)

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Step-by-step explanation: