Answer :

Let us draw AF, perpendicular to BC
such that AF is the height of \(\triangle \)ABD, \(\triangle \)ADE and \(\triangle \)AEC

\(\therefore \) area of triangle =( \(\frac{1}{2} \)) × base ×Height

\(\therefore \) Area of \(\triangle \) ABD= (\(\frac{1}{2} \) ) × BD × AF

\(\therefore \) Area of \(\triangle \) ADE = (\(\frac{1}{2} \) ) × DE × AF

\(\therefore \) Area of \(\triangle \) AEC = (\(\frac{1}{2} \) ) × EC × AF

Since , BD= DE= EC (Given )

\(\therefore \) Area of \(\triangle \) ABD= Area of \(\triangle \) ADE= Area of \(\triangle \) AEC

- Parallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle.
- In figure, ABCD, DCFE and ABFE are parallelograms. Show that ar(ADE) = ar(BCF).
- In figure, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, show that ar(\(\triangle \)BPC) = ar(\(\triangle \)DPQ).[Hint Join AC.]
- In figure, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, Show that (i) ar (BDE) = \(\frac{1}{4} \) ar (ABC) (ii) ar (BDE) = \(\frac{1}{2} \) ar (BAE) (iii) ar (ABC) = 2 ar (BEC) (iv) ar (BFE) = ar (AFD) (v) ar (BFE) = 2 ar (FED) (vi) ar (FED) = \(\frac{1}{8} \) ar (AFC) [Hint : Join EC and AD. Show that BE || AC and DE || AB, etc.]
- Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that ar(APB) x ar(CPD) = ar(APD) x ar(BPC). [Hint From A and C, draw perpendiculars to BD.]
- P and Q are respectively the mid-points of sides AB and BC of a triangle ABC and R is the mid-point of AP, show that (i) ar(PRQ) = \(\frac{1}{2} \) ar(ARC) (ii) ar(PBQ) = ar(ARC) (iii) ar(RQC) = \(\frac{3}{8} \) ar(ABC)
- In figure, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX \(\perp \) DE meets BC at Y. Show that (i) \(\triangle \)MBC \(\cong \) \(\triangle \)ABD (ii) ar(BYXD) = 2 ar(MBC) (iii) ar(BYXD) = ax(ABMN) (iv) \(\triangle \)FCB \(\cong \) \(\triangle \)ACE (v) ar(CYXE) = 2 ar(FCB) (vi) ar(CYXE) = ar(ACFG) (vii) ar(BCED) = ar(ABMN) + ar(ACFG)

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