# (1) D and E are points on sides AB and AC respectively of triangle ABC such that ar (DBC) = ar (EBC). Prove that DE//BC(2) Diagnols 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(3) Diagnols AC and BD of a quadrilateral ABCD intersect each other at P. Show that ar (APB) x ar (CPD) = ar (APD) x ar (BPC)CAN SOMEONE PLZZZZ GIVE THE ANSWERS TO THESE QUESTIONS .vryy imp...

Q 1

Given that-

D and E are points on sides AB and AC respectively of ΔABC Also, area (DBC) = area (EBC)

We know that Area of Any triangle is Since, Area of ΔDBC and ΔEBC are same and Base is common for both the triangles

⇒ height of ΔDBC is also equal to the height of ΔBEC

h1 = h2

Hence   BC || DE

Q 2. Given that:-

Diagonals of triangle ABCD intersects at O and

area (ΔAOD) = area (BOC)

⇒ area (ΔAOD) + area (ΔAOB) = area (ΔBOC) + area (ΔAOB)

⇒ area (ΔABD) = area (ΔACB)

Again, Base of these both triangles is same

⇒ height h1 and h2 of the both triangles is also same.

⇒ DD' = CC'

⇒ AB || DC

⇒ ABCD is a trapezium.

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