In the figure given below , ABCD is a parallelogram and P is any point in BC.Prove that ,area of triangle ABP area of triangle DPC = area of triangle APD
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Hello Manisha
Please check https://www.teachoo.com/1235/448/Ex-9.2--4---In-figure--P-is-a-point-in-the-interior-of/category/Ex-9.2/
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Hello Manisha
Please check https://www.teachoo.com/1235/448/Ex-9.2--4---In-figure--P-is-a-point-in-the-interior-of/category/Ex-9.2/
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