(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

⇒ *h*_{1} = *h*_{2}

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 *h*_{1} and *h*_{2} of the both triangles is also same.

⇒ DD' = CC'

⇒ AB || DC

⇒ ABCD is a trapezium.

**Que.3**Your question is not clear, check it again and then get back to us so that we can give you a meaningful help

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