Please give me the best way for solving the problem of class 9^{th} ncert math of Areas of Parallelograms and Triangles chapter of math of class 9^{th} of exercise 9.4 of question no 5(3) what is the tricky way for solving this question In Fig.9.33, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that:(iii) ar (ABC) = 2 ar (BEC)

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# In Fig.9.33, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that:(iii) ar (ABC) = 2 ar (BEC) Q.5(3)

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ar(Î”ABE) = ar(Î”BEC) [Common base BE and BE || AC]

ar(Î”ABF) + ar(Î”BEF) = ar(Î”BEC)

From eq

^{n}(i), we get,ar(Î”ABF) + ar(Î”AFD) = ar(Î”BEC)

ar(Î”ABD) = ar(Î”BEC)

Â½ ar(Î”ABC) = ar(Î”BEC)

ar(Î”ABC) = 2 ar(Î”BEC)

Hence proved