Verified answer

Appropriate Question:

In Fig., P is a point in the interior of a parallelogram ABCD. Show that

(i) ar (APB) + ar (PCD) = 1/2 ar (ABCD)

(ii) ar (APD) + ar (PBC)=ar (APB) + ar (PCD)

[tex]\large\underline{\sf{Solution-}}[/tex]

Through P, draw a line EF parallel to AB intersecting AD at E and BC at F.

So, We have AB || CD || EF

Now, Parallelogram EFCD and triangle PCD are on the same base and in between parallel lines CD and EF.

[tex]\implies\sf\:\dfrac{1}{2} Area( { \parallel}^{gm} \: EFCD) = Area( \triangle \: PCD) - - - (1) \\ [/tex]

Again, Parallelogram ABEF and triangle PAB are on the same base and in between parallel lines CD and EF.

[tex]\implies\sf\:\dfrac{1}{2} Area( { \parallel}^{gm} \: ABEF) = Area( \triangle \: PAB) - - - (2) \\ [/tex]

On adding equation (1) and (2), we get

[tex]\sf\:Area( \triangle \: PCD) + Area( \triangle \: PAB) = \dfrac{1}{2} Area( { \parallel}^{gm} \: EFCD) + \dfrac{1}{2} Area( { \parallel}^{gm} \: ABEF) \\ [/tex]

[tex]\sf\:Area( \triangle \: PCD) + Area( \triangle \: PAB) = \dfrac{1}{2}\left[ Area( { \parallel}^{gm} \: EFCD) + Area( { \parallel}^{gm} \: ABEF)\right] \\ [/tex]

[tex]\implies\bf\:Area( \triangle \: APB) + Area( \triangle \: PCD) = \dfrac{1}{2}Area( { \parallel}^{gm} \: ABCD) - - (i) \\ [/tex]

Now, Through P, draw a line GH parallel to BC intersecting AB at G and CD at H.

So, We have CD || BC || GH

Now, Parallelogram ADHG and triangle APD are on the same base and in between parallel lines AD and GH.

[tex]\implies\sf\:\dfrac{1}{2} Area( { \parallel}^{gm} \: ADHG) = Area( \triangle \: PAD) - - - (3) \\ [/tex]

Again, Parallelogram BCHG and triangle PCD are on the same base and in between parallel lines BC and GH.

[tex]\implies\sf\:\dfrac{1}{2} Area( { \parallel}^{gm} \: BCHG) = Area( \triangle \: PBC) - - - (4) \\ [/tex]

On adding equation (3) and (4), we get

[tex]\sf\:Area( \triangle \: APD) + Area( \triangle \: PBC) = \dfrac{1}{2} Area( { \parallel}^{gm} \: ADHG) + \dfrac{1}{2} Area( { \parallel}^{gm} \: BCHG) \\ [/tex]

[tex]\sf\:Area( \triangle \: APD) + Area( \triangle \: PBC) = \dfrac{1}{2}\left[ Area( { \parallel}^{gm} \: ADHG) + Area( { \parallel}^{gm} \: BCHG)\right] \\ [/tex]

[tex]\implies\sf\:Area( \triangle \: APD) + Area( \triangle \: PBC) = \dfrac{1}{2}\ Area( { \parallel}^{gm} \: ABCD) - - (ii) \\ [/tex]

So, from equation (i) and (ii), we concluded that

[tex]\implies\bf\:Area( \triangle \: APD) + Area( \triangle \: PBC) = Area( \triangle \: APB) + Area( \triangle \: PCD) \\ [/tex]