Question

Que : Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that

ar (APB) × ar (CPD) = ar (APD) × ar (BPC).

Answer 1

Answer:

Solution:

Given: The diagonal AC and BD of the quadrilateral ABCD, intersect each other at point E.

Construction:

From A, draw AM perpendicular to BD

From C, draw CN perpendicular to BD

To Prove: ar(ΔAED) ar(ΔBEC) = ar (ΔABE) ar (ΔCDE)

Proof:

ar(ΔABE) = ½ ×BE×AM………….. (i)

ar(ΔAED) = ½ ×DE×AM………….. (ii)

Dividing eq. (ii) by (i) , we get,

ar(ΔAED)/ar(ΔABE) = [1/2×DE×AM]/1/2×BE×Am]

= DE/BE ………..(iii)

Similarly,

ar(ΔCDE)/ar(ΔBEC) = DE/BE …….(iv)

From eq. (iii) and (iv), we get;

ar(ΔAED)/ar(ΔABE) = ar(ΔCDE)/ar(ΔBEC)

= ar(ΔAED) × ar(ΔBEC) = ar (ΔABE) × ar (ΔCDE)

Hence proved.

Answer 2

Step-by-step explanation:

Let ABCD is a quadrilateral having diagonal AC and BD. These diagonals intersect at point P.

Now area of a triangle = 1/2 * base * height

=> Area(ΔAPB) * Area(ΔCPD) = {1/2 * BP*AM}*{1/2 * PD*CN}

=> Area(ΔAPB) * Area(ΔCPD) = 1/4 * BP*AM* PD*CN .........1

Again

Area(ΔAPD) * Area(ΔBPC) = {1/2 * PD*AM}*{1/2 *CN*BP}

=> Area(ΔAPD) * Area(ΔBPC) = 1/4 * BP*AM* PD*CN ..........2

From equation 1 and 2, we get

Area(ΔAPB) * Area(ΔCPD) = Area(ΔAPD) * Area(ΔBPC)