ABCD is a square and EF is parallel to diagonal BD and EM=FM. prove that

DF=BE

AM bisects angle BAD

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

∴     CBD =   CDB = 90 2 = 45 " src="https://s3mn.mnimgs.com/img/shared/editlive_temp/mathmlequation8289848564628255736.png" style="margin: 0px; padding: 0px; border: 0px; max-width: 550px;" /

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

(ii)∆ABE ≅ ∆ADF (by SAS congruency criterion)

⇒ BAE = DAF and AE = AF ........(3)

and ∆AER ≅ ∆AFR (by SSS congruency criterion)

⇒ EAR = FAR .........(4)

Now Adding (3) and (4) we get

⇒ BAE + EAR = DAF + FAR

⇒ BAR = DAR

i.e AR bisects BAD

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In🔼AME and 🔼AMF Angle AME = Angle AMF = 90° (given) EM=FM (given) AM=AM (common) ➡🔼AME ~= 🔼AMF (SAS criteria) ➡AE=AF (CPCT)......... (i) ➡angle EAM= angleFAM (CPCT)........ (ii) Now, In 🔼ADF and 🔼ABE AD=AB (sides of a square) Angle D= Angle B=90° (angles of a square) AF=AE (from (i)) ➡🔼ADF~=🔼ABE (R.H.S criteria) ➡DF=BE or BE=DF (CPCT) ➡angle BAE= angleDAF (CPCT)..... (iii) Now from (ii) and (iii) ➡angle BAE+angleEAM =angle DAF +angle FAM ➡angle BAM =angle DAM ➡AM bisects angle BAD PROVED PROVED
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in the given figure ABCD is a square and EF is parallel to diagonal DB and EM=FM prove BF=DE AM bisects angle BAD
 
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