P is any point on diagonal AC of parallelogram ABCD. Show that area of triangle ADP is equal to area of triangle ABP
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in ΔADP and ΔABP
∠APB= ∠APD (each 90 degree)
AP=AP (common side)
BP=DP (we know that diagonal of a parallelogram bisect each other)
∴ΔADP≅ΔABP By Side Angle Side congruency
and we know that if two triangles are congruent then there area's are also same
area ΔADP=area ΔABP
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