Question

In the given figure diagonals PR and QS of quadrilateral PQRS intersect each other at A. show that ar(∆PSA) × ar(∆QAR) = ar(∆PAQ) × ar(∆SAR)......solve it on a paper and plz post the pic of the solution

Answer 1

Hey buddy your answer is in the picture.

Hope it helps.

Have a good day dear.

Answer 2

Answer:

Step-by-step explanation:

Draw SN⊥PR and QN⊥PR.

therefore, area ΔPSA×areaΔQAR=$(\frac{1}{2}PA{\times}SM)(\frac{1}{2}AR{\times}QN)$

=$\frac{1}{4}{\times}PM{\times}SM{\times}AR{\times}QN.....$                 (1)

Also, area ΔSAR×areaΔPAQ=$(\frac{1}{2}{\times}AR{\times}SM)(\frac{1}{2}PA{\times}QN)$

=$\frac{1}{4}QA{\times}PM{\times}SA{\times}PN$                                   (2)

Now, from equation (1) and (2), we get

area ΔPSA×areaΔQAR=area ΔSAR×areaΔPAQ

Hence proved.