Question

In the given figure, abcd is a square in which E is the mid-point of side BC. if AF=BE=CG, find the angles <CGE, <FEG and show that BC||FG.​

Answer 1

Answer:

given

ABCD is a square

AF = BE = CG

Also E is the mid point of BC

proof:

Since E is the mid point of BC

Therefore, BE = CE

      Also BE = CG

             CE = CG

In triangle GCE

GC = CE

angle CGE = angle CEG

Since ABCD is a square

Therefore  , each angle is equal to 90°-------> 1

By ASP of a triangle

∠ECG+∠CGE+∠GEC = 180°

∠ECG = 90°[from '1']

90°+2∠CGE=180°

2∠CGE = 90°

∠CGE = 45°

In triangle GCE and FBE

∠ECG = ∠EBF [ proved above ]

CE = BE [ E is the mid point of BC]

∠FEB=∠GEC

Therefore by ASA rule triangle GCE ≅FBE

by CPCT BF = CG

     

∠CEB = 180°

∠CEG+∠FEG+∠FEB=180°

∠FEG=180°-[45+45]

∠FEG=90°

Now

    since AF = CG

      also CG = FB [ proved above]

         AF = FB

          therefore F is the mid point of BA------>2

Now

     AF + BF = CG + DG

     AF = CG+DG-BF

   AF = DG

Therefore G is the mid-point of DC---------->3

Therefore BC||FG.​