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:
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