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प्रश्न
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.
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उत्तर
It is given that
Area (ΔAOD) = Area (ΔBOC)
Area (ΔAOD) + Area (ΔAOB) = Area (ΔBOC) + Area (ΔAOB)
Area (ΔADB) = Area (ΔACB)
We know that triangles on the same base having areas equal to each other lie between the same parallels.
Therefore, these triangles, ΔADB and ΔACB, are lying between the same parallels.
i.e., AB || CD
Therefore, ABCD is a trapezium.
संबंधित प्रश्न
Show that the diagonals of a parallelogram divide it into four triangles of equal area.
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.
(ii) ar (DEF) = 1/4ar (ABC)
(iii) ar (BDEF) = 1/2ar (ABC)
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.]
D and E are points on sides AB and AC respectively of ΔABC such that
ar (DBC) = ar (EBC). Prove that DE || BC.
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.]
In the below fig. D and E are two points on BC such that BD = DE = EC. Show that ar
(ΔABD) = ar (ΔADE) = ar (ΔAEC).
ABCD is a parallelogram and X is the mid-point of AB. If ar (AXCD) = 24 cm2, then ar (ABC) = 24 cm2.
In the following figure, ABCD and EFGD are two parallelograms and G is the mid-point of CD. Then ar (DPC) = `1/2` ar (EFGD).
X and Y are points on the side LN of the triangle LMN such that LX = XY = YN. Through X, a line is drawn parallel to LM to meet MN at Z (See figure). Prove that ar (LZY) = ar (MZYX)
If the medians of a ∆ABC intersect at G, show that ar (AGB) = ar (AGC) = ar (BGC) = `1/3` ar (ABC)
