Question

ABCD is a square. The bisector of ∠DBC cuts AC and CD at E and F respectively. Prove that BF × CE = BE × DF.

Answer 1

Answer:

A/q

BF is bisector of angDBC

i.e. angDBF = angCBF

we know each angle of a square is 90° which is bisected by its diagonals hence becomes 45° now as BF is intersecting angle DBC the angles are now 22.5 degree each. also angle BDC = BCE = 45°

•°• BDF~BCE [AA similarly]

by one theorem we also know that the ratio of the areas of two similar triangle is equal to the ratio of the squares of their corresponding sides.

so, (DF/CE)^2 = (BF/BE)^2 => DF/CE= BF/BE= DF*BE=BF*CE       proved

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