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show that ar (ABE) = ar (ACF)

Answer :

XY || BC = EY || BC ...(Given)

Also, BE || AC = BE || CY

\(\therefore\) we can say that, EBYC is a parallelogram.

Similarly,

XY || BC = XF || BC

Also, FC || AB = FC || XB

\(\therefore \) BCFX is a parallelogram.

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

\(\therefore \) Area (EBCY) = (\(\frac{1}{2} \) ) Area (BCFX) ...(i)

Now, considering parallelogram EBYC and \(\triangle{AEB}\),

We can say that, these lie on the same base BE and are between the same parallels BE and AC.

\(\therefore \) Area (\(\triangle{AEB}\)) = (\(\frac{1}{2} \) ) Area (EBYC) ...(ii)

Also, parallelogram BCFX and \(\triangle{ACF}\) are on the same base CF and between the same parallels CF and AB.

Thus, Area (\(\triangle{ACF}\)) = (\(\frac{1}{2} \) ) Area (BCFX) ...(iii)

So, from eq.(i), (ii) and (iii), we get,

Area (\(\triangle{AEB}\)) = (\(\triangle{ACF}\))

Hence, proved.

- In Figure, E is any point on median AD of a \(\triangle{ABC}\). Show that ar (ABE) = ar (ACE)
- In a triangle ABC, E is the mid-point of median AD. Show that ar (BED) = \(\frac{1}{4} \) ar (ABC).
- Show that the diagonals of a parallelogram divide it into four triangles of equal area.
- In Figure, ABC and ABD are two triangles on the same base AB. If line segment CD is bisected by AB at O, show that ar(ABC) = ar (ABD).
- D, E and F are respectively the mid-points of the sides BC, CA and AB of a \(\triangle{ABC}\). Show thati) BDEF is a parallelogram.ii) ar(DEF) = \(\frac{1}{4} \) ar(ABC) iii) ar(BDEF) = \(\frac{1}{2} \) ar(ABC)
- In Figure, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that:i) ar (DOC) = ar (AOB)ii) ar (DCB) = ar (ACB)iii) DA || CB or ABCD is a parallelogram.[Hint: From D and B, draw perpendiculars to AC.]
- D and E are points on sides AB and AC respectively of \(\triangle{ABC}\) such that ar (DBC) = ar (EBC). Prove that DE || BC.
- 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 Figure). Show that ar (ABCD) = ar (PBQR).[Hint : Join AC and PQ. Now compare ar (ACQ) and ar (APQ).]
- Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar (AOD) = ar (BOC).
- In Figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show thati) ar (ACB) = ar (ACF)ii) ar (AEDF) = ar (ABCDE).
- A villager Itwaari has a plot of land of the shape of a quadrilateral. The Gram Panchayat of the village decided to take over some portion of his plot from one of the corners to construct a Health Centre. Itwaari agrees to the above proposal with the condition that he should be given equal amount of land in lieu of his land adjoining his plot so as to form a triangular plot. Explain how this proposal will be implemented.
- ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar (ADX) = ar (ACY).[Hint: Join CX.]
- In Figure, AP || BQ || CR. Prove that ar (AQC) = ar (PBR).
- Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar (AOD) = ar (BOC). Prove that ABCD is a trapezium.
- In Figure, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.

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