Question

In the figure ABCD is a square and ef is parallel to the diagonal bd and EM=FM. Prove that a) DF=BE b) 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.

Answer 2

explanation is in the photo