Proof: Let ABCD be a parallelogram and AC be a diagonal (see Fig. 8.2). Observe that the diagonal AC divides parallelogram ABCD into two triangles, namely, ΔABC and ΔCDA.
In ΔABC and ΔCDA, 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, ΔABC ≅ ΔCDA (ASA rule)
or, diagonal AC divides parallelogram ABCD into two congruent triangles ABC and CDA.
Proof: You have already proved that a diagonal divides the parallelogram into two congruent triangles; so the corresponding parts, the corresponding sides, are equal.
So, AB = DC and AD = BC
Proof: Let sides AB and CD of the quadrilateral ABCD be equal and also AD = BC (see Fig. 8.3). Draw diagonal AC.
Clearly, ΔABC ≅ ΔCDA (Why?)
So, ∠BAC = ∠DCA and ∠BCA = ∠DAC (Why?)
Therefore, ABCD is a parallelogram.
Proof: Draw a parallelogram and measure its angles. Each pair of opposite angles is equal.
Proof: Draw a parallelogram ABCD and draw both its diagonals intersecting at the point O (see Fig. 8.4).
Measure the lengths of OA, OB, OC and OD.
You will observe that OA = OC and OB = OD.
or, O is the mid-point of both the diagonals.
Proof: Note that in Fig. 8.5, it is given that OA = OC and OB = OD.
So, ΔAOB ≅ ΔCOD (Why?)
Therefore, ∠ABO = ∠CDO (Why?)
From this, we get AB || CD
Similarly, BC || AD
Therefore ABCD is a parallelogram.
Proof: Observe Fig 8.16 in which E and F are mid-points of AB and AC respectively and CD || BA.
ΔAEF ≅ ΔCDF (ASA Rule)
So, EF = DF and BE = AE = DC (Why?)
Therefore, BCDE is a parallelogram. (Why?)
This gives EF || BC.
In this case, also note that EF = ½ ED = ½ BC.
Proof: In Fig 8.17, observe that E is the mid-point of AB, line l is passing through E and is parallel to BC and CM || BA.
Prove that AF = CF by using the congruence of ΔAEF and ΔCDF.
Solution: Let ABCD be a rectangle in which ∠A = 90°.
We have to show that ∠B = ∠C = ∠D = 90°
We have, AD || BC and AB is a transversal (see Fig. 8.6).
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 parallelogram)
So, ∠C = 90° and ∠D = 90°.
Therefore, each of the angles of a rectangle is a right angle.
Solution: Consider the rhombus ABCD (see Fig. 8.7).
You know that AB = BC = CD = DA (Why?)
Now, in ΔAOD and ΔCOD,
OA = OC (Diagonals of a parallelogram bisect each other)
OD = OD (Common)
AD = CD
Therefore, ΔAOD ≅ ΔCOD (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.
Solution:
(i) ΔABC 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.
Solution: It is given that PS || QR and transversal p intersects them at points A and C respectively.
The bisectors of ∠PAC and ∠ACQ intersect at B and bisectors of ∠ACR and ∠SAC intersect at D.
We are to show that quadrilateral ABCD is a rectangle.
Now, ∠PAC = ∠ACR (Alternate angles as l || m and p is a transversal)
So, ½ ∠PAC = ½ ∠ACR
i.e., ∠BAC = ∠ACD
These form a pair of alternate angles for lines AB and DC with AC as transversal and they are equal also.
So, AB || DC
Similarly, BC || AD (Considering ∠ACB and ∠CAD)
Therefore, quadrilateral ABCD is a parallelogram.
Also, ∠PAC + ∠CAS = 180° (Linear pair)
So, ½ ∠PAC + ½ ∠CAS = ½ × 180° = 90°
or, ∠BAC + ∠CAD = 90°
or, ∠BAD = 90°
So, ABCD is a parallelogram in which one angle is 90°.
Therefore, ABCD is a rectangle.
Solution: Let P, Q, R and S be the points of intersection of the bisectors of ∠A and ∠B, ∠B and ∠C, ∠C and ∠D, and ∠D and ∠A respectively of parallelogram ABCD (see Fig. 8.10).
In ΔASD, since DS bisects ∠D and AS bisects ∠A, therefore,
∠DAS + ∠ADS = ½ ∠A + ½ ∠D = ½ (∠A + ∠D) = ½ × 180° = 90°
(∠A and ∠D are interior angles on the same side of the transversal)
Also, ∠DAS + ∠ADS + ∠DSA = 180° (Angle sum property of a triangle)
or, 90° + ∠DSA = 180°
or, ∠DSA = 90°
So, ∠PSR = 90° (Being vertically opposite to ∠DSA)
Similarly, it can be shown that ∠APB = 90° or ∠SPQ = 90° (as it was shown for ∠DSA). Similarly, ∠PQR = 90° and ∠SRQ = 90°.
So, PQRS is a quadrilateral in which all angles are right angles.
We have shown that ∠PSR = ∠PQR = 90° and ∠SPQ = ∠SRQ = 90°. So both pairs of opposite angles are equal.
Therefore, PQRS is a parallelogram in which one angle (in fact all angles) is 90° and so, PQRS is a rectangle.
Solution: As D and E are mid-points of sides AB and BC of the triangle ABC, by Theorem 8.8, 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, ΔBDE ≅ ΔFED
Similarly ΔDAF ≅ ΔFED and ΔEFC ≅ ΔFED
So, all the four triangles are congruent.
Solution: We are given that AB = BC and have to prove that DE = EF.
Let us join A to F intersecting m at G.
The trapezium ACFD is divided into two triangles; namely ΔACF and ΔAFD.
In ΔACF, 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 Theorem 8.9)
Now, in ΔAFD, we can apply the same argument as G is the mid-point of AF, GE || AD and so by Theorem 8.9, 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.
Solution: Let ABCD be a parallelogram with AC = BD.
In ΔABC and ΔDCB:
AB = DC (Opposite sides of parallelogram)
BC = BC (Common)
AC = BD (Given)
So, ΔABC ≅ ΔDCB (SSS congruence)
Therefore, ∠ABC = ∠DCB (CPCT)
But ∠ABC + ∠DCB = 180° (Adjacent angles of parallelogram)
So, 2∠ABC = 180° ⇒ ∠ABC = 90°
Similarly, all angles are 90°.
Hence, ABCD is a rectangle.
Solution: Let ABCD be a square.
In ΔABC and ΔDCB:
AB = DC (Sides of square)
BC = BC (Common)
∠ABC = ∠DCB = 90° (Angles of square)
So, ΔABC ≅ ΔDCB (SAS congruence)
Therefore, AC = BD (CPCT)
Hence, diagonals are equal.
Also, in square, diagonals bisect each other (property of parallelogram).
In ΔAOB and ΔBOC:
AO = OC (Diagonals bisect each other)
BO = BO (Common)
AB = BC (Sides of square)
So, ΔAOB ≅ ΔBOC (SSS congruence)
Therefore, ∠AOB = ∠BOC (CPCT)
But ∠AOB + ∠BOC = 180° (Linear pair)
So, 2∠AOB = 180° ⇒ ∠AOB = 90°
Hence, diagonals bisect each other at right angles.
Solution:
(i) Since AB || DC and AC is transversal,
∠BAC = ∠DCA (Alternate interior angles) ...(1)
Since AD || BC and AC is transversal,
∠DAC = ∠BCA (Alternate interior angles) ...(2)
But ∠BAC = ∠DAC (Given, AC bisects ∠A) ...(3)
From (1), (2) and (3):
∠DCA = ∠BCA
So, AC bisects ∠C.
(ii) In ΔABC and ΔADC:
∠BAC = ∠DAC (Given)
∠BCA = ∠DCA (Proved above)
AC = AC (Common)
So, ΔABC ≅ ΔADC (ASA congruence)
Therefore, AB = AD and CB = CD (CPCT)
But in parallelogram ABCD, AB = CD and AD = BC
So, AB = BC = CD = DA
Hence, ABCD is a rhombus.
Solution:
(i) Since AC bisects ∠A and ∠C,
∠DAC = ∠BAC and ∠DCA = ∠BCA
In rectangle ABCD, ∠A = ∠C = 90°
So, ∠DAC = ∠BAC = 45° and ∠DCA = ∠BCA = 45°
In ΔABC, ∠ABC = 90° (Angle of rectangle)
So, ∠BAC + ∠BCA + ∠ABC = 180°
45° + 45° + ∠ABC = 180° ⇒ ∠ABC = 90°
Similarly, all angles are 90°.
In ΔABC and ΔADC:
∠ABC = ∠ADC = 90°
∠BAC = ∠DAC = 45°
AC = AC (Common)
So, ΔABC ≅ ΔADC (AAS congruence)
Therefore, AB = AD (CPCT)
But in rectangle, AB = CD and AD = BC
So, AB = BC = CD = DA
Hence, ABCD is a square.
(ii) In square, diagonals bisect the angles.
So, BD bisects ∠B and ∠D.
Solution:
(i) In ΔAPD and ΔCQB:
AD = CB (Opposite sides of parallelogram)
∠ADP = ∠CBQ (Alternate interior angles, AD || BC)
DP = BQ (Given)
So, ΔAPD ≅ ΔCQB (SAS congruence)
(ii) AP = CQ (CPCT from (i))
(iii) In ΔAQB and ΔCPD:
AB = CD (Opposite sides of parallelogram)
∠ABQ = ∠CDP (Alternate interior angles, AB || CD)
BQ = DP (Given)
So, ΔAQB ≅ ΔCPD (SAS congruence)
(iv) AQ = CP (CPCT from (iii))
(v) Since AP = CQ and AQ = CP, and opposite sides are equal,
APCQ is a parallelogram.
Solution:
(i) In ΔAPB and ΔCQD:
AB = CD (Opposite sides of parallelogram)
∠ABP = ∠CDQ (Alternate interior angles, AB || CD)
∠APB = ∠CQD = 90° (Given, AP ⊥ BD and CQ ⊥ BD)
So, ΔAPB ≅ ΔCQD (AAS congruence)
(ii) AP = CQ (CPCT from (i))
Solution: [Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]
Draw CE || DA intersecting AB produced at E.
Since AB || CD and CE || DA, AECD is a parallelogram.
So, AD = CE and AE = CD
But AD = BC (Given)
So, BC = CE
In ΔBCE, BC = CE ⇒ ∠CBE = ∠CEB
But ∠CEB = ∠DAB (Corresponding angles, CE || DA)
And ∠CBE = ∠CBA (Same angle)
So, ∠DAB = ∠CBA ⇒ ∠A = ∠B
Similarly, we can prove ∠C = ∠D
In ΔABC and ΔBAD:
AB = BA (Common)
BC = AD (Given)
∠ABC = ∠BAD (Proved above)
So, ΔABC ≅ ΔBAD (SAS congruence)
Therefore, AC = BD (CPCT)
Solution:
(i) In ΔADC, S and R are mid-points of AD and CD respectively.
So, by mid-point theorem, SR || AC and SR = ½ AC
(ii) Similarly, in ΔABC, P and Q are mid-points of AB and BC respectively.
So, PQ || AC and PQ = ½ AC
Therefore, PQ = SR
(iii) Since PQ || AC and SR || AC, so PQ || SR
Also, PQ = SR
So, PQRS is a parallelogram.
Solution: Join AC and BD.
In ΔABC, P and Q are mid-points of AB and BC respectively.
So, PQ || AC and PQ = ½ AC
Similarly, in ΔADC, SR || AC and SR = ½ AC
So, PQ || SR and PQ = SR
Therefore, PQRS is a parallelogram.
Now, in ΔABD, P and S are mid-points of AB and AD respectively.
So, PS || BD
But AC ⊥ BD (Diagonals of rhombus are perpendicular)
Since PS || BD and PQ || AC, and AC ⊥ BD,
So, PS ⊥ PQ ⇒ ∠SPQ = 90°
Therefore, PQRS is a rectangle.
Solution: Join AC and BD.
In ΔABC, P and Q are mid-points of AB and BC respectively.
So, PQ || AC and PQ = ½ AC
Similarly, in ΔADC, SR || AC and SR = ½ AC
So, PQ || SR and PQ = SR
Similarly, PS || QR and PS = QR
Therefore, PQRS is a parallelogram.
Now, in rectangle ABCD, AC = BD (Diagonals of rectangle are equal)
So, PQ = ½ AC = ½ BD = PS
Therefore, PQ = PS
So, adjacent sides of parallelogram PQRS are equal.
Hence, PQRS is a rhombus.
Solution: Let diagonal BD intersect EF at G.
In ΔABD, E is mid-point of AD and EG || AB
So, by converse of mid-point theorem, G is mid-point of BD.
In ΔBCD, G is mid-point of BD and GF || DC (since AB || DC and EF || AB)
So, by converse of mid-point theorem, F is mid-point of BC.
Solution: Since ABCD is a parallelogram, AB || CD and AB = CD
E and F are mid-points of AB and CD respectively.
So, AE = ½ AB = ½ CD = CF
Also, AE || CF (since AB || CD)
So, AECF is a parallelogram.
In ΔABP, E is mid-point of AB and EQ || AP (since AF || EC)
So, by converse of mid-point theorem, Q is mid-point of BP.
Similarly, in ΔDQC, F is mid-point of DC and FP || CQ (since AF || EC)
So, P is mid-point of DQ.
Let BP = x, then PQ = x (since Q is mid-point of BP)
Let PD = y, then PQ = y (since P is mid-point of DQ)
So, x = y
Therefore, BP = PQ = QD
Hence, AF and EC trisect the diagonal BD.
Solution: Let ABCD be a quadrilateral. P, Q, R, S are mid-points of AB, BC, CD, DA respectively.
Join PQ, QR, RS, SP and also join PR and QS.
In ΔABC, P and Q are mid-points of AB and BC respectively.
So, PQ || AC and PQ = ½ AC
Similarly, in ΔADC, S and R are mid-points of AD and CD respectively.
So, SR || AC and SR = ½ AC
Therefore, PQ || SR and PQ = SR
So, PQRS is a parallelogram.
In a parallelogram, diagonals bisect each other.
So, PR and QS bisect each other.
Hence, the line segments joining the mid-points of opposite sides of a quadrilateral bisect each other.
Solution:
(i) In ΔABC, M is mid-point of AB and MD || BC
So, by converse of mid-point theorem, D is mid-point of AC.
(ii) Since MD || BC and AC is transversal,
∠ADM = ∠ACB (Corresponding angles)
But ∠ACB = 90° (Given)
So, ∠ADM = 90° ⇒ MD ⊥ AC
(iii) In ΔADM and ΔCDM:
AD = CD (D is mid-point of AC)
MD = MD (Common)
∠ADM = ∠CDM = 90°
So, ΔADM ≅ ΔCDM (SAS congruence)
Therefore, AM = CM (CPCT)
But AM = ½ AB (M is mid-point of AB)
So, CM = MA = ½ AB