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.

Since diagonal of square bisects the angles.

So, ∠CBD = ∠CDB = 45°

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.