Solution - Triangles on the Same Base and Between the Same Parallels



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ConceptTriangles on the Same Base and Between the Same Parallels  


Show that the diagonals of a parallelogram divide it into four triangles of equal area.


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In the following 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) = 1/4 ar (ABC)

(ii) ar (BDE) = 1/2 ar (BAE)

(iii) ar (ABC) = 2 ar (BEC)

(iv) ar (BFE) = ar (AFD)

(v) ar (BFE) = 2 ar (FED)

(vi) ar (FED) = 1/8 ar (AFC)

[Hint : Join EC and AD. Show that BE || AC and DE || AB, etc.]

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In the given 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.]

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Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that ar (APB) × ar (CPD) = ar (APD) × ar (BPC).

[Hint : From A and C, draw perpendiculars to BD.]

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In a triangle ABC, E is the mid-point of median AD. Show that ar (BED) = 1/4ar (ABC).

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D, E and F are respectively the mid-points of the sides BC, CA and AB of a ΔABC. Show that

(i) BDEF is a parallelogram.

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