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!

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