Example-1 :- Show that each angle of a rectangle is a right angle.
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-2 :- Show that the diagonals of a rhombus are perpendicular to each other.
Solution :-
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.
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.
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 :-
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.
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.
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.
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.
