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