Mathematics
In the given figure; AD is median of △ ABC and E is any point on median AD. Prove that Area (△ ABE) = Area (△ ACE).
Theorems on Area
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Answer
We know that,
Median of a triangle divides it into two triangles of equal area.
Given,
AD is the median of △ ABC.
∴ Area of △ ABD = Area of △ ACD ……….(1)
Since, E is a point on median AD.
∴ ED is median of △ EBC.
∴ Area of △ EBD = Area of △ ECD ……….(2)
Subtracting equation (2) from (1), we get :
⇒ Area of △ ABD - Area of △ EBD = Area of △ ACD - Area of △ ECD
⇒ Area of △ ABE = Area of △ ACE.
Hence, proved that area (△ ABE) = area (△ ACE).
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Related Questions
In the following figure, DE is parallel to BC. Show that :
(i) Area (△ ADC) = Area (△ AEB)
(ii) Area (△ BOD) = Area (△ COE)
Show that :
(i) a diagonal divides a parallelogram into two triangles of equal area.
(ii) the ratio of the areas of two triangles of the same height is equal to the ratio of their bases.
(iii) the ratio of the areas of two triangles on the same base is equal to the ratio of their heights.
In the figure of question 14, if E is the mid point of median AD, then prove that :
Area (△ ABE) = Area (△ ABC).
ABCD is a parallelogram. P and Q are the mid-points of sides AB and AD respectively. Prove that area of triangle APQ = of the area of parallelogram ABCD.