If each diagonal of a quadrilateral seperates it into two triangles of equal area, prove that the quadrilateral is a parallelogram.
Dear Student!
Here is the answer to your query.
Given : A quadrilateral ABCD and its diagonals AC and BD divides it into two triangles of equal areas i.e.
ar (ΔABD) = ar (ΔCDB)
and ar (ΔABC) = ar (ΔACD)
To prove : ABCD is a parallelogram
Proof : ar (ΔABC) = ar (ΔACD) ..........(1)
ar (ΔABC) + ar (ΔACD) = ar (quad ABCD) ...........(2)
from (1) and (2)
2 ar (ΔABC) = ar (quad ΔABCD) ......(3)
and ar (ΔABD) = ar (ΔBCD) .............(4)
ar (ΔABD) + ar (ΔBCD) = ar (quad ABCD) .......... (5)
from (4) and (5)
2 ar (ΔABD) = ar (quad ΔABCD) ..........(6)
from (3) and (6) we get
2 ar (ΔABC) = 2 ar (ΔABD)
⇒ ar (ΔABC) = ar (ΔABD)
Since ΔABC and ΔABD are on the some base AB. Therefore they must have equal corresponding altitudes.
i.e. Altitude from C of ΔABC = Altitude from D of ΔABD
⇒ DC || AB
Similarly AD || BC
Hence ABCD is a parallelogram.
Cheers!