Please guide me the best way for solving the question of class 9th math of Areas of Parallelograms and Triangles chapter of exercise 9.4 of math of question no. 6 What is the best way for solving this question, please guide me Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that ar (APB)×ar (CPD) = ar (APD)×ar (BPC). [Hint : From A and C, draw perpendiculars to BD.]
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Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that ar (APB)×ar (CPD) = ar (APD)×ar (BPC). [Hint : From A and C, draw perpendiculars to BD.] Q.6
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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(AED)/ar(ABE) = DE/BE…….. (iii)
Similarly,
ar(CDE)/ar(BEC) = DE/BE ……. (iv)
From eq. (iii) and (iv) , we get
ar(AED)/ar(ABE) = ar(CDE)/ar(BEC)
, ar(ΔAED)×ar(ΔBEC) = ar(ΔABE)×ar (ΔCDE)
Hence proved.