Given: P is any point on the median AD of a ΔABC
We know, median of a triangle divides it into two triangles of equal area.
Therefore, ar (ΔADB) = ar (ΔADC) ...(1)
Also, ar (ΔPDB) = ar (ΔPDC) ...(2)
Subtracting (2) from (1), we get
ar (ΔADB) − ar (ΔPDB) = ar (ΔADC) − ar (ΔPDC)
⇒ ar (ΔABP) = ar (ΔACP)
⇒= 1
Hence, = 1.