4exam

In the given figure, P is a point in the interior of a parallelogram ABCD. Show that (i) ar (APB) + ar (PCD) =   ar (ABCD) (ii) ar (APD) + ar (PBC) = ar (APB) + ar (PCD) [Hint: Through. P, draw a line parallel to AB] Consider ΔACD. Line-segment CD is bisected by AB at O. Therefore, AO is the median of ΔACD. ∴ Area (ΔACO) = Area (ΔADO) ... (1) Considering ΔBCD, BO is the median. ∴ Area (ΔBCO) = Area (ΔBDO) ... (2) Adding equations (1) and (2), we obtain Area (ΔACO) + Area (ΔBCO) = Area (ΔADO) + Area (ΔBDO) ⇒ Area (ΔABC) = Area (ΔABD)

In the given figure, E is any point on median AD of a ΔABC. Show that ar (ABE) = ar (ACE) ans) AD is the median of ΔABC. Therefore, it will divide ΔABC into two triangles of equal areas.  ∴ Area (ΔABD) = Area (ΔACD) ... (1) ED is the median of ΔEBC. ∴ Area (ΔEBD) = Area (ΔECD) ... (2) On subtracting equation (2) from equation (1), we obtain Area (ΔABD) − Area (EBD) = Area (ΔACD) − Area (ΔECD) Area (ΔABE) = Area (ΔACE)

We know that diagonals of parallelogram bisect each other. Therefore, O is the mid-point of AC and BD. BO is the median in ΔABC. Therefore, it will divide it into two triangles of equal areas. ∴ Area (ΔAOB) = Area (ΔBOC) ... (1) In ΔBCD, CO is the median. ∴ Area (ΔBOC) = Area (ΔCOD) ... (2) Similarly, Area (ΔCOD) = Area (ΔAOD) ... (3) From equations (1), (2), and (3), we obtain Area (ΔAOB) = Area (ΔBOC) = Area (ΔCOD) = Area (ΔAOD) Therefore, it is evident that the diagonals of a parallelogram divide it into four triangles of equal area.

AD is the median of ΔABC. Therefore, it will divide ΔABC into two triangles of equal areas. ∴ Area (ΔABD) = Area (ΔACD) ⇒  ... (1) In ΔABD, E is the mid-point of AD. Therefore, BE is the median. ∴ Area (ΔBED) = Area (ΔABE) ⇒ Area (ΔBED) =  Area (ΔABD) ⇒ Area (ΔBED) =  Area (ΔABC) [From equation (1)] ⇒ Area (ΔBED) =  Area (ΔABC)