TOPICS
Unit-9(Theorems)

Theorem-1 :-  A diagonal of a parallelogram divides it into two congruent triangles.

Solution :-
        parallelogram
  Given that : ABCD is a Parallelogram and AC be diagonal. 
  Prove that : Δ ABC ≅ Δ CDA .
  Proof : The diagonal AC divides parallelogram ABCD into two triangles, namely, Δ ABC and Δ CDA. 
          In Δ ABC and Δ CDA, note that BC || AD and AC is a transversal. 
          So, ∠ BCA = ∠ DAC (Pair of alternate angles)
          Also, AB || DC and AC is a transversal. 
          So, ∠ BAC = ∠ DCA (Pair of alternate angles) and AC = CA (Common) 
          So, Δ ABC ≅ Δ CDA (ASA rule).
    

Theorem-2 :-  In a parallelogram, opposite sides are equal.

Solution :-
        parallelogram
  Given that : ABCD is a Parallelogram and AC be diagonal.   
  Prove that : AB = CD and AD = BC.
  Proof : The diagonal AC divides parallelogram ABCD into two triangles, namely, Δ ABC and Δ CDA. 
          In Δ ABC and Δ CDA, note that BC || AD and AC is a transversal. 
          So, ∠ BCA = ∠ DAC (Pair of alternate angles)
          Also, AB || DC and AC is a transversal. 
          So, ∠ BAC = ∠ DCA (Pair of alternate angles) and AC = CA (Common) 
          So, Δ ABC ≅ Δ CDA (ASA rule).
          By CPCT rule, AB = CD and AD = BC.
    

Theorem-3 :-  If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram.

Solution :-
        parallelogram
  Given that : Let sides AB and CD of the quadrilateral ABCD be equal and also  AD = BC.  
  Prove that :  ABCD is a parallelogram.
  Construction : Draw diagonal AC.
  Proof : In Δ ABC and Δ CDA, note that BC || AD and AC is a transversal. 
          So, ∠ BCA = ∠ DAC (Pair of alternate angles)
          Also, AB || DC and AC is a transversal. 
          So, ∠ BAC = ∠ DCA (Pair of alternate angles) and AC = CA (Common) 
          So, Δ ABC ≅ Δ CDA (ASA rule).
          By CPCT rule, AB = CD and AD = BC.
          So, ∠ BAC = ∠ DCA and ∠ BCA = ∠ DA.
          Hence, ABCD is a Parallelogram.
    

Theorem-4 :-  In a parallelogram, opposite angles are equal.

Solution :-
        parallelogram
  Given that : ABCD is a Parallelogram. 
  Prove that : ∠ DAB = ∠ BCD and ∠ ADC = ∠ ABC.
  Construction : Draw AC be diagonal.
  Proof : The diagonal AC divides parallelogram ABCD into two triangles, namely, Δ ABC and Δ CDA. 
          In Δ ABC and Δ CDA, note that BC || AD and AC is a transversal. 
          So, ∠ BCA = ∠ DAC (Pair of alternate angles)
          Also, AB || DC and AC is a transversal. 
          So, ∠ BAC = ∠ DCA (Pair of alternate angles) and AC = CA (Common) 
          So, Δ ABC ≅ Δ CDA (ASA rule).
          By CPCT rule, ∠ BAC = ∠ DCA ....(i) and 
          By CPCT rule, ∠ DAC = ∠ BCA ....(ii)
          Adding eq(i) and eq(ii)
          ∠ BAC + ∠ DAC = ∠ DCA + ∠ BCA
          ∠ DAB = ∠ BCD        
         Similarly, ∠ ADC = ∠ ABC.
    

Theorem-5 :-  The diagonals of a parallelogram bisect each other.

Solution :-
        parallelogram
  Given that : ABCD is a Parallelogram. AC and BD are be diagonal whoose intersect point on O.   
  Prove that : OA = OC and OB = OD.
  Proof : In Δ AOD and Δ BOC,
          ∠ BCO = ∠ DAO (Pair of alternate angles)
          ∠ CBO = ∠ ADO (Pair of alternate angles) and 
          BC = DA (parallelogram sides are equal) 
          So, Δ AOD ≅ Δ BOC (ASA rule).
          By CPCT rule, OB = OD and OA = OC.
    

Theorem-6 :- : If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.

Solution :-
        parallelogram
  Given that : ABCD is a Parallelogram. AC and BD are be diagonal whoose intersect point on O. 
               AO = OC, and OB = OD.  
  Prove that : ABCD is a Parallelogram.
  Proof : In Δ AOD and Δ BOC,
          AO = OC (Given)
          OB = OD (Given) and 
          ∠BOC = ∠DOA (parallelogram sides are equal) 
          So, Δ AOD ≅ Δ BOC (SAS rule).
          By CPCT rule, ∠ BCO = ∠ DAO  and ∠ CBO = ∠ ADO.
          So, AB||CD and BC||AD.
          Hence, ABCD is a Parallelogram.
    

Theorem-7 :-  The line segment joining the mid-points of two sides of a triangle is parallel to the third side.

Solution :-
        parallelogram
  Given that : In figure E and F are mid-points of AB and AC respectively and CD || BA.
               So, AF = CF.
  Prove that : EF || BC.
  Proof : In Δ AEF and Δ CDF 
          ∠ EAF = ∠ DCF (Pair of alternate angles)
          ∠ AEF = ∠ CDF (Pair of alternate angles) and 
          AF = CF (Given) 
          So, Δ AEF ≅ Δ CDF (ASA rule).
          By CPCT rule, EF = DF and BE = AE = DC 
          Therefore, BCDE is a parallelogram. 
          This gives EF || BC.
    
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