Q. 334.4( 8 Votes )

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.


Answer :

Since diagonal of square bisects the angles.

So, CBD = CDB = 45°  [ Also all angles of square are right angles i.e. half of all is 45°]   (1)


Also similarly ABD = ADB = 45°


Since lines EF || BD


By corresponding angles-


CEF = CDB = 45°


Also CFE = CBD = 45°


So, CE = CF {since sides opposite to equal angles are equal} …(i)


And CD = BC {sides of a square are equal} …(ii)


Subtracting I from II


CD-CE = BC-CF


So, BF = DE


Also let’s consider ΔADX and ΔABX {where X is intersection point of AM and BD}


ABD = ADB = 45°


AX is a common side.


AD = AB {sides of a square are equal}


The triangles are congruent by SAS (side angle side) criteria.


So, DAM = MAB (congruency criteria)


Hence AM bisects BAD.

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