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Question

In the given figure, ABCD is a square and EF is parallel to diagonal DB and EM = FM. Prove that: (i) BF = DE (ii) AM bisects ∠BAD.
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Solution

i We have,CD = BC-------1 Sides of a square are equal and BDC = DBC = 45°Given: EFDB BDC = FEC Corresponding angles FEC = 45°Similarly, EFC = 45°CEF is an isosceles triangle.and CE = CF -------2From 1 - 2, we get: CD - CE = BC - CF DE = BFBF = DE

ii In ABF and ADE,AB= AD All sides of a square are equalABF = ADE Both angles are equal to 90°BF = DE (Proved above) ABF ADE by SAS congruency criteriaBAF = DAE ----3And AF = AE ------4Now, In AME and AMF,AM = AM Common sideEM = FM GivenAE = AF from 4 AME AMF By SSS congruency criteriaEAM = FAM FAM = EAM -----5From 3 + 5, we get:BAF+FAM = DAE+EAMBAM = DAMTherefore, AM bisects BAD.

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