TOPICS
Unit-9(Examples)

Example-1 :-  Show that each angle of a rectangle is a right angle.

Solution :-
        Rectangle
  Given that : Let ABCD be a rectangle in which ∠ A = 90°. 
  Prove that :  ∠ B = ∠ C = ∠ D = 90°
  Proof : We have, AD || BC and AB is a transversal. 
          So, ∠ A + ∠ B = 180° (Interior angles on the same side of the transversal) 
          But, ∠ A = 90° So, ∠ B = 180° – ∠ A = 180° – 90° = 90° 
          Now, ∠ C = ∠ A and ∠ D = ∠ B (Opposite angles of the parallellogram) 
          So, ∠ C = 90° and ∠ D = 90°. 
          Therefore, each of the angles of a rectangle is a right angle. 
    

Example-2 :-  Show that the diagonals of a rhombus are perpendicular to each other.

Solution :-
        Rhombus
  Given that : ABCD is a Rhombus and AB = BC = CD = AD.   
  Prove that : The diagonals of a rhombus are perpendicular to each other.
  Proof : Now, in Δ AOD and Δ COD, 
          OA = OC (Diagonals of a parallelogram bisect each other) 
          OD = OD (Common) 
          AD = CD (Given)
          Therefore, Δ AOD ≅ Δ COD (SSS congruence rule) 
          This gives, ∠ AOD = ∠ COD (CPCT) 
          But, ∠ AOD + ∠ COD = 180° (Linear pair) 
          So, 2∠ AOD = 180° or, 
          ∠ AOD = 90° 
          So, the diagonals of a rhombus are perpendicular to each other.
    

Example-3 :- : ABC is an isosceles triangle in which AB = AC. AD bisects exterior angle PAC and CD || AB. Show that
(i) ∠ DAC = ∠ BCA and (ii) ABCD is a parallelogram. Rhombus

Solution :-
  Given that : Δ ABC is isosceles. So, AB = AC.
  Prove that : (i) ∠ DAC = ∠ BCA and (ii) ABCD is a parallelogram. 
  Proof :  (i) Δ ABC is isosceles in which AB = AC (Given) 
           So, ∠ ABC = ∠ ACB (Angles opposite to equal sides) 
           Also, ∠ PAC = ∠ ABC + ∠ ACB (Exterior angle of a triangle) or, 
           ∠ PAC = 2∠ ACB .....(1) 
           Now, AD bisects ∠ PAC. 
           So, ∠ PAC = 2∠ DAC .....(2) 
           Therefore, 2∠ DAC = 2∠ ACB [From (1) and (2)] or, 
           ∠ DAC = ∠ ACB 
        
           (ii) Now, these equal angles form a pair of alternate angles when line segments 
           BC and AD are intersected by a transversal AC. 
           So, BC || AD 
           Also, BA || CD (Given) 
           Now, both pairs of opposite sides of quadrilateral ABCD are parallel. 
           So, ABCD is a parallelogram.
    

Example-4 :-  Two parallel lines l and m are intersected by a transversal p. Show that the quadrilateral formed by the bisectors of interior angles is a rectangle. Rectangle

Solution :-
  Given that : Let ABCD be a rectangle in which ∠ A = 90°. 
  Prove that :  ∠ B = ∠ C = ∠ D = 90°
  Proof : We have, AD || BC and AB is a transversal. 
          So, ∠ A + ∠ B = 180° (Interior angles on the same side of the transversal) 
          But, ∠ A = 90° So, ∠ B = 180° – ∠ A = 180° – 90° = 90° 
          Now, ∠ C = ∠ A and ∠ D = ∠ B (Opposite angles of the parallellogram) 
          So, ∠ C = 90° and ∠ D = 90°. 
          Therefore, each of the angles of a rectangle is a right angle. 
    

Example-5 :- Show that the bisectors of angles of a parallelogram form a rectangle.

Solution :-
        Rhombus
  Given that : ABCD is a Rhombus and AB = BC = CD = AD.   
  Prove that : The diagonals of a rhombus are perpendicular to each other.
  Proof : Now, in Δ AOD and Δ COD, 
          OA = OC (Diagonals of a parallelogram bisect each other) 
          OD = OD (Common) 
          AD = CD (Given)
          Therefore, Δ AOD ≅ Δ COD (SSS congruence rule) 
          This gives, ∠ AOD = ∠ COD (CPCT) 
          But, ∠ AOD + ∠ COD = 180° (Linear pair) 
          So, 2∠ AOD = 180° or, 
          ∠ AOD = 90° 
          So, the diagonals of a rhombus are perpendicular to each other.
    

Example-6 :- : ABCD is a parallelogram in which P and Q are mid-points of opposite sides AB and CD. If AQ intersects DP at S and BQ intersects CP at R, show that:
(i) APCQ is a parallelogram.
(ii) DPBQ is a parallelogram.
(iii) PSQR is a parallelogram. parallelogram

Solution :-
  Given that : ABCD is a parallelogram in which P and Q are mid-points of opposite sides AB and CD.
  Prove that : (i) APCQ is a parallelogram, (ii) DPBQ is a parallelogram and (iii) PSQR is a parallelogram. 
  Proof : (i) In quadrilateral APCQ, AP || QC (Since AB || CD ).....(1)
          AP = 1/2 x AB, CQ = 1/2 x CD (Given)
          Also, AB = CD   
          So, AP = QC ......(2) 
          Therefore, APCQ is a parallelogram [From (1) and (2) and Theorem] 
          
          (ii) Similarly, quadrilateral DPBQ is a parallelogram, 
          because DQ || PB and DQ = PB.
        
          (iii) In quadrilateral PSQR, 
          SP || QR (SP is a part of DP and QR is a part of QB) 
          Similarly, SQ || PR
          So, PSQR is a parallelogram.
    

Example-7 :-  In Δ ABC, D, E and F are respectively the mid-points of sides AB, BC and CA. Show that Δ ABC is divided into four congruent triangles by joining D, E and F. triangle

Solution :-
  Given that : D and E are mid-points of sides AB and BC of the Δ ABC
  Prove that : Δ ABC is divided into four congruent triangles by joining D, E and F. 
  Proof : Given that D and E are mid-points of sides AB and BC of the Δ ABC, 
          By mid point Theorem, 
          DE || AC 
          Similarly, 
          DF || BC and EF || AB 
          Therefore ADEF, BDFE and DFCE are all parallelograms. 
          Now DE is a diagonal of the parallelogram BDFE, 
          Therefore, Δ BDE ≅ Δ FED 
          Similarly Δ DAF ≅ Δ FED 
          and Δ EFC ≅ Δ FED 
          So, all the four triangles are congruent.
    

Example-8 :- l, m and n are three parallel lines intersected by transversals p and q such that l, m and n cut off equal intercepts AB and BC on p. Show that l, m and n cut off equal intercepts DE and EF on q also. trapezium

Solution :-
  Given that :  AB = BC    
  Prove that :  DE = EF.
  Construction : Let us join A to F intersecting m at G..
  Proof : The trapezium ACFD is divided into two triangles;
          i.e., Δ ACF and Δ AFD. 
          In Δ ACF, it is given that B is the mid-point of AC (AB = BC) and BG || CF (since m || n).
          So, G is the mid-point of AF (by using mid point Theorem) 
          Now, in Δ AFD, we can apply the same argument as G is the mid-point of AF, GE || AD and 
          so By mid point Theorem, E is the mid-point of DF, 
          i.e., DE = EF. 
          In other words, l, m and n cut off equal intercepts on q also.
    
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