Question

ABCD is square and EF parallel to DBand EM=FM .Prove that (1)BF=DE (2)AM bisects angle BAD

Answer 1

Solution:-

(1) Since diagonal of a square bisects the vertex and BD is the diagonal of square ABCD.

∴ ∠ CBD = ∠ CDB = 90/2 = 45°

Given :  EF || BD

⇒ ∠ CEF = ∠ CBD = 45° and ∠ CEF = ∠ CDB = 45° (Corresponding angles)

⇒ CEF = CFE

⇒ CE = CF (Sides opposite of equal angles are equal) .....(1)

Now, BC = CD (Sides of square) .....(2)

Subtracting (1) from (2), we get

⇒ BC CE = CD CF

⇒ BE = DF or DF = BE    (First condition proved)

(2) Δ ABE ≡ ADF (By SAS congruency criterion)

⇒ ∠ BAE = ∠ DAF .....(3)

 AE = AF

And, Δ AEM ≡ Δ AFM (By SSS congruency criterion)

⇒ ∠ EAM = ∠ FAM ....(4)

Now adding (3) and (4), we get

⇒ BAE + EAM = DAF + FAM

⇒ BAM = DAM

i.e. AM bisects ∠ BAD 

Proved.