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प्रश्न
In the following figure, CD || AE and CY || BA. Prove that ar (CBX) = ar (AXY).
उत्तर
Given: In the following figure, CD || AE and CY || BA
To prove: ar (ΔCBX) = ar (ΔAXY) .
Proof: We know that, triangles on the same base and between the same parallels are equal in areas.
Here, ΔABY and ΔABC both lie on the same base AB and between the same parallels CY and BA.
ar (ΔABY) = ar (ΔABC)
⇒ ar (ABX) + ar (AXY) = ar (ABX) + ar (CBX)
⇒ ar (AXY) = ar (CBX) ...[Eliminating ar (ABX) from both sides]
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संबंधित प्रश्न
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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.
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(i) ar (ACB) = ar (ACF)
(ii) ar (AEDF) = ar (ABCDE)
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