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

**Solution:**

If a triangle and parallelogram are lying on the same base and between the same parallel lines, the area of the triangle will be equal to half of the area of a parallelogram.

Also, if two parallelograms are lying on the same base and between the same pair of parallel lines then both of them will have equal area.

Let’s draw points X and Y, intersected by line EF on sides AB and AC respectively.

Let's consider BCYE

It is given that, XY || BC so, EY || BC

Also, BE || AC so, BE || CY

Therefore, BCYE is a parallelogram.

Similarly, In BCFX

It is given that, XY || BC so, XF || BC

Since, CF || AB so, CF || BX

Therefore, BCFX is a parallelogram.

Parallelograms BCYE and BCFX are lying on the same base BC and between the same parallels BC and EF.

Therefore, According to Theorem 9.1: Parallelograms on the same base and between the same parallels are equal in area.

∴ Area (BCYE) = Area (BCFX) ... (1)

Now, Consider parallelogram BCYE and ΔAEB

They are lying on the same base BE and are between the same parallels BE and AC.

∴ Area (ΔABE) = 1/2 Area (BCYE)... (2)

Also, parallelogram BCFX and ΔACF are lying on the same base CF and existing between the same parallels CF and AB.

∴ Area (ΔACF) = 1/2 Area (BCFX)... (3)

From Equations (1), (2), and (3), we obtain

Area (ΔABE) = Area (ΔACF) proved.

**☛ Check: **NCERT Solutions for Class 9 Maths Chapter 9

**Video Solution:**

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

Maths NCERT Solutions Class 9 Chapter 9 Exercise 9.3 Question 8

**Summary:**

If XY is a line parallel to side BC of triangle ABC, and BE || AC, CF || AB meet XY at E and F, then ar (ABE) = ar (ACF).

**☛ Related Questions:**

- The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed (see Fig. 9.26). Show that ar (ABCD) = ar (PBQR).
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