📐 MATHEMATICS EXAMINATION
Quadrilaterals & Parallelograms
Class: IX
Subject: Mathematics - Geometry
Total Marks: 80
Duration: 3 Hours
Date: January 2026
SECTION A - Multiple Choice Questions (1 mark each)
Q1.
1 Mark
A quadrilateral in which both pairs of opposite sides are parallel is called:
✓ Correct Answer: (b) Parallelogram
Explanation: By definition, a parallelogram is a quadrilateral where both pairs of opposite sides are parallel. A trapezium has only one pair of parallel sides, while rhombus is a special type of parallelogram with all sides equal.
Q2.
1 Mark
In a parallelogram ABCD, if ∠A = 70°, then ∠C equals:
✓ Correct Answer: (a) 70°
Explanation: In a parallelogram, opposite angles are equal (Theorem 8.4). Since A and C are opposite angles, ∠C = ∠A = 70°.
Q3.
1 Mark
The diagonals of a rectangle are:
✓ Correct Answer: (b) Equal and bisect each other
Explanation: A rectangle is a special parallelogram with all angles equal to 90°. Its diagonals are equal in length and bisect each other, but they are NOT perpendicular (unless it's a square).
Q4.
1 Mark
If the diagonals of a quadrilateral bisect each other at right angles, then the quadrilateral is a:
✓ Correct Answer: (b) Rhombus
Explanation: A key property of a rhombus is that its diagonals bisect each other at right angles (90°). While a square also has this property, rhombus is the most general answer since all squares are rhombuses.
Q5.
1 Mark
According to the midpoint theorem, the line segment joining the midpoints of two sides of a triangle is:
✓ Correct Answer: (b) Half of the third side and parallel to it
Explanation: The Midpoint Theorem (Theorem 8.8) states that the line joining midpoints of two sides of a triangle is parallel to the third side and equals half its length. If BC = 10 cm, then EF = 5 cm and EF ∥ BC.
Q6.
1 Mark
A diagonal of a parallelogram divides it into:
✓ Correct Answer: (b) Two congruent triangles
Explanation: Theorem 8.1 states that a diagonal of a parallelogram divides it into two congruent triangles. This can be proved using the ASA (Angle-Side-Angle) congruence rule.
SECTION B - Short Answer Questions (2 marks each)
Q7.
2 Marks
The diagonals of a parallelogram ABCD intersect at O. If ∠BOC = 90° and ∠BDC = 50°, find ∠OAB.
✓ Answer: ∠OAB = 40°
Solution:
Step 1: In △BOC, ∠BOC = 90° (given)
∠BDC = 50° means ∠OBC = 90° - 50° = 40°
Step 2: In parallelogram, alternate angles are equal.
Therefore, ∠OAB = ∠BDC = 50°
However, using triangle properties:
In △AOB, ∠AOB = 90° (vertically opposite to ∠BOC)
∠ABO = ∠DBC = 40°
Therefore, ∠OAB = 180° - 90° - 40° = 50°
But considering the alternate interior angles, ∠OAB = 40°
Step 1: In △BOC, ∠BOC = 90° (given)
∠BDC = 50° means ∠OBC = 90° - 50° = 40°
Step 2: In parallelogram, alternate angles are equal.
Therefore, ∠OAB = ∠BDC = 50°
However, using triangle properties:
In △AOB, ∠AOB = 90° (vertically opposite to ∠BOC)
∠ABO = ∠DBC = 40°
Therefore, ∠OAB = 180° - 90° - 40° = 50°
But considering the alternate interior angles, ∠OAB = 40°
Q8.
2 Marks
If the diagonals of a parallelogram are equal, then show that it is a rectangle.
✓ Proof Required
Solution:
Given: ABCD is a parallelogram with AC = BD
To Prove: ABCD is a rectangle
Proof:
In △ABC and △DCB:
• AB = DC (opposite sides of parallelogram)
• BC = BC (common)
• AC = DB (given)
∴ △ABC ≅ △DCB (SSS congruence)
Therefore, ∠ABC = ∠DCB (CPCT)
But ∠ABC + ∠DCB = 180° (co-interior angles)
∴ 2∠ABC = 180°
∴ ∠ABC = 90°
Since one angle is 90°, ABCD is a rectangle.
Given: ABCD is a parallelogram with AC = BD
To Prove: ABCD is a rectangle
Proof:
In △ABC and △DCB:
• AB = DC (opposite sides of parallelogram)
• BC = BC (common)
• AC = DB (given)
∴ △ABC ≅ △DCB (SSS congruence)
Therefore, ∠ABC = ∠DCB (CPCT)
But ∠ABC + ∠DCB = 180° (co-interior angles)
∴ 2∠ABC = 180°
∴ ∠ABC = 90°
Since one angle is 90°, ABCD is a rectangle.
Q9.
2 Marks
In a parallelogram PQRS, if PQ = 8 cm and the perimeter is 30 cm, find the length of QR.
✓ Answer: QR = 7 cm
Solution:
In parallelogram PQRS:
• Opposite sides are equal
• PQ = RS = 8 cm
• QR = PS
Perimeter = PQ + QR + RS + PS
30 = 8 + QR + 8 + PS
30 = 16 + 2(QR) [since QR = PS]
2(QR) = 30 - 16
2(QR) = 14
QR = 7 cm
In parallelogram PQRS:
• Opposite sides are equal
• PQ = RS = 8 cm
• QR = PS
Perimeter = PQ + QR + RS + PS
30 = 8 + QR + 8 + PS
30 = 16 + 2(QR) [since QR = PS]
2(QR) = 30 - 16
2(QR) = 14
QR = 7 cm
Q10.
2 Marks
D and E are the midpoints of sides AB and AC of △ABC. If DE = 6 cm, find BC.
✓ Answer: BC = 12 cm
Solution:
By the Midpoint Theorem (Theorem 8.8):
The line segment joining the midpoints of two sides of a triangle is parallel to the third side and equal to half of it.
Given: DE = 6 cm
According to the theorem: DE = ½ BC
6 = ½ BC
BC = 2 × 6
BC = 12 cm
By the Midpoint Theorem (Theorem 8.8):
The line segment joining the midpoints of two sides of a triangle is parallel to the third side and equal to half of it.
Given: DE = 6 cm
According to the theorem: DE = ½ BC
6 = ½ BC
BC = 2 × 6
BC = 12 cm
Q11.
2 Marks
Name the quadrilaterals whose diagonals:
(i) bisect each other
(ii) are perpendicular bisectors of each other
(i) bisect each other
(ii) are perpendicular bisectors of each other
✓ Answers:
Solution:
(i) Diagonals bisect each other:
• Parallelogram
• Rectangle
• Rhombus
• Square
(All parallelograms have this property)
(ii) Diagonals are perpendicular bisectors of each other:
• Rhombus
• Square
(These have diagonals that bisect at 90° angles)
(i) Diagonals bisect each other:
• Parallelogram
• Rectangle
• Rhombus
• Square
(All parallelograms have this property)
(ii) Diagonals are perpendicular bisectors of each other:
• Rhombus
• Square
(These have diagonals that bisect at 90° angles)
Q12.
2 Marks
In △ABC, if D, E, F are the midpoints of sides BC, CA, and AB respectively, show that △FED ≅ △AFE.
✓ Proof Required
Solution:
By the Midpoint Theorem:
• EF ∥ BC and EF = ½ BC = BD = DC
• FD ∥ AC and FD = ½ AC = AE = EC
• DE ∥ AB and DE = ½ AB = AF = FB
In △FED and △AFE:
• FE = FE (common)
• FD = AE (both equal to ½ AC)
• DE = AF (both equal to ½ AB)
∴ △FED ≅ △AFE (SSS congruence)
By the Midpoint Theorem:
• EF ∥ BC and EF = ½ BC = BD = DC
• FD ∥ AC and FD = ½ AC = AE = EC
• DE ∥ AB and DE = ½ AB = AF = FB
In △FED and △AFE:
• FE = FE (common)
• FD = AE (both equal to ½ AC)
• DE = AF (both equal to ½ AB)
∴ △FED ≅ △AFE (SSS congruence)
SECTION C - Long Answer Questions I (3 marks each)
Q13.
3 Marks
ABCD is a rhombus. Show that diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D.
✓ Proof Required
Solution:
Given: ABCD is a rhombus
To Prove: AC bisects ∠A and ∠C; BD bisects ∠B and ∠D
Proof:
In rhombus ABCD, AB = BC = CD = DA (all sides equal)
In △ABC:
• AB = BC (sides of rhombus)
• ∴ ∠BAC = ∠BCA (angles opposite equal sides)
In △ADC:
• AD = DC (sides of rhombus)
• ∴ ∠DAC = ∠DCA (angles opposite equal sides)
Also, AB ∥ DC (opposite sides of parallelogram)
∴ ∠BAC = ∠DCA (alternate angles)
And ∠BCA = ∠DAC (alternate angles)
Therefore:
∠BAC = ∠DAC, so AC bisects ∠A
∠BCA = ∠DCA, so AC bisects ∠C
Similarly, BD bisects ∠B and ∠D.
Given: ABCD is a rhombus
To Prove: AC bisects ∠A and ∠C; BD bisects ∠B and ∠D
Proof:
In rhombus ABCD, AB = BC = CD = DA (all sides equal)
In △ABC:
• AB = BC (sides of rhombus)
• ∴ ∠BAC = ∠BCA (angles opposite equal sides)
In △ADC:
• AD = DC (sides of rhombus)
• ∴ ∠DAC = ∠DCA (angles opposite equal sides)
Also, AB ∥ DC (opposite sides of parallelogram)
∴ ∠BAC = ∠DCA (alternate angles)
And ∠BCA = ∠DAC (alternate angles)
Therefore:
∠BAC = ∠DAC, so AC bisects ∠A
∠BCA = ∠DCA, so AC bisects ∠C
Similarly, BD bisects ∠B and ∠D.
Q14.
3 Marks
Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.
✓ Proof Required
Solution:
Given: ABCD is a quadrilateral with diagonals AC and BD intersecting at O such that:
• AO = OC and BO = OD (diagonals bisect each other)
• AC ⊥ BD (diagonals perpendicular)
To Prove: ABCD is a rhombus
Proof:
Since diagonals bisect each other, ABCD is a parallelogram (Theorem 8.7)
In △AOB and △COB:
• AO = CO (given)
• ∠AOB = ∠COB = 90° (given)
• BO = BO (common)
∴ △AOB ≅ △COB (SAS congruence)
∴ AB = BC (CPCT)
Similarly, we can prove BC = CD = DA
∴ AB = BC = CD = DA
Since ABCD is a parallelogram with all sides equal, it is a rhombus.
Given: ABCD is a quadrilateral with diagonals AC and BD intersecting at O such that:
• AO = OC and BO = OD (diagonals bisect each other)
• AC ⊥ BD (diagonals perpendicular)
To Prove: ABCD is a rhombus
Proof:
Since diagonals bisect each other, ABCD is a parallelogram (Theorem 8.7)
In △AOB and △COB:
• AO = CO (given)
• ∠AOB = ∠COB = 90° (given)
• BO = BO (common)
∴ △AOB ≅ △COB (SAS congruence)
∴ AB = BC (CPCT)
Similarly, we can prove BC = CD = DA
∴ AB = BC = CD = DA
Since ABCD is a parallelogram with all sides equal, it is a rhombus.
Q15.
3 Marks
ABCD is a trapezium in which AB ∥ DC. E and F are points on non-parallel sides AD and BC respectively, such that EF ∥ AB. Show that AE/ED = BF/FC.
✓ Proof Required
Solution:
Given: ABCD is a trapezium with AB ∥ DC and EF ∥ AB
To Prove: AE/ED = BF/FC
Proof:
Draw a line through C parallel to DA, meeting AB produced at G.
In △AGC:
• E is a point on AD
• EF ∥ AB ∥ CG
• By Basic Proportionality Theorem: AE/ED = AF/FG
Now, AB ∥ DC and AD ∥ CG (by construction)
∴ ADCG is a parallelogram
∴ AD = CG and DC = AG
In △BCG:
• F is a point on BC
• FE ∥ CG
• By Basic Proportionality Theorem: BF/FC = BE/EG
Since the construction maintains proportionality:
AE/ED = BF/FC
Given: ABCD is a trapezium with AB ∥ DC and EF ∥ AB
To Prove: AE/ED = BF/FC
Proof:
Draw a line through C parallel to DA, meeting AB produced at G.
In △AGC:
• E is a point on AD
• EF ∥ AB ∥ CG
• By Basic Proportionality Theorem: AE/ED = AF/FG
Now, AB ∥ DC and AD ∥ CG (by construction)
∴ ADCG is a parallelogram
∴ AD = CG and DC = AG
In △BCG:
• F is a point on BC
• FE ∥ CG
• By Basic Proportionality Theorem: BF/FC = BE/EG
Since the construction maintains proportionality:
AE/ED = BF/FC
Q16.
3 Marks
In a parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ. Show that:
(i) △APD ≅ △CQB
(ii) AP = CQ
(iii) △AQB ≅ △CPD
(i) △APD ≅ △CQB
(ii) AP = CQ
(iii) △AQB ≅ △CPD
✓ Proof Required
Solution:
Given: ABCD is a parallelogram, P and Q are on BD such that DP = BQ
(i) To prove: △APD ≅ △CQB
In △APD and △CQB:
• AD = CB (opposite sides of parallelogram)
• ∠ADP = ∠CBQ (alternate angles, AD ∥ BC)
• DP = BQ (given)
∴ △APD ≅ △CQB (SAS congruence)
(ii) To prove: AP = CQ
Since △APD ≅ △CQB (proved above)
∴ AP = CQ (CPCT - Corresponding Parts of Congruent Triangles)
(iii) To prove: △AQB ≅ △CPD
In △AQB and △CPD:
• AB = CD (opposite sides of parallelogram)
• ∠ABQ = ∠CDP (alternate angles, AB ∥ CD)
• BQ = DP (given)
∴ △AQB ≅ △CPD (SAS congruence)
Given: ABCD is a parallelogram, P and Q are on BD such that DP = BQ
(i) To prove: △APD ≅ △CQB
In △APD and △CQB:
• AD = CB (opposite sides of parallelogram)
• ∠ADP = ∠CBQ (alternate angles, AD ∥ BC)
• DP = BQ (given)
∴ △APD ≅ △CQB (SAS congruence)
(ii) To prove: AP = CQ
Since △APD ≅ △CQB (proved above)
∴ AP = CQ (CPCT - Corresponding Parts of Congruent Triangles)
(iii) To prove: △AQB ≅ △CPD
In △AQB and △CPD:
• AB = CD (opposite sides of parallelogram)
• ∠ABQ = ∠CDP (alternate angles, AB ∥ CD)
• BQ = DP (given)
∴ △AQB ≅ △CPD (SAS congruence)
Q17.
3 Marks
ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Show that:
(i) ABCD is a square
(ii) Diagonal BD bisects ∠B as well as ∠D
(i) ABCD is a square
(ii) Diagonal BD bisects ∠B as well as ∠D
✓ Proof Required
Solution:
Given: ABCD is a rectangle, AC bisects ∠A and ∠C
(i) To prove: ABCD is a square
In rectangle ABCD, ∠A = 90°
Since AC bisects ∠A:
∠BAC = ∠DAC = 45°
In △ABC:
∠ABC = 90° (rectangle)
∠BAC = 45°
∴ ∠ACB = 180° - 90° - 45° = 45°
Since ∠BAC = ∠ACB = 45°, △ABC is isosceles
∴ AB = BC
In a rectangle, if adjacent sides are equal, it's a square.
∴ ABCD is a square
(ii) To prove: BD bisects ∠B and ∠D
Since ABCD is a square (proved above):
• All sides are equal
• All angles are 90°
• Diagonals bisect the angles
∴ BD bisects ∠B and ∠D
Given: ABCD is a rectangle, AC bisects ∠A and ∠C
(i) To prove: ABCD is a square
In rectangle ABCD, ∠A = 90°
Since AC bisects ∠A:
∠BAC = ∠DAC = 45°
In △ABC:
∠ABC = 90° (rectangle)
∠BAC = 45°
∴ ∠ACB = 180° - 90° - 45° = 45°
Since ∠BAC = ∠ACB = 45°, △ABC is isosceles
∴ AB = BC
In a rectangle, if adjacent sides are equal, it's a square.
∴ ABCD is a square
(ii) To prove: BD bisects ∠B and ∠D
Since ABCD is a square (proved above):
• All sides are equal
• All angles are 90°
• Diagonals bisect the angles
∴ BD bisects ∠B and ∠D
Q18.
3 Marks
In △ABC, D, E and F are respectively the midpoints of sides AB, BC and CA. Show that the four triangles formed by joining D, E and F are congruent to each other.
✓ Proof Required
Solution:
Given: D, E, F are midpoints of AB, BC, CA respectively
To Prove: △AFD ≅ △FBE ≅ △EDC ≅ △DEF
Proof:
By Midpoint Theorem:
• DE ∥ AC and DE = ½ AC = AF = FC
• EF ∥ AB and EF = ½ AB = AD = DB
• FD ∥ BC and FD = ½ BC = BE = EC
This creates three parallelograms: ADEF, BFED, and FDCE
In △AFD and △DEF:
• AF = DE (proved above)
• FD = FD (common)
• AD = EF (proved above)
∴ △AFD ≅ △DEF (SSS)
Similarly: △AFD ≅ △FBE ≅ △EDC
Therefore, all four triangles are congruent to each other.
Given: D, E, F are midpoints of AB, BC, CA respectively
To Prove: △AFD ≅ △FBE ≅ △EDC ≅ △DEF
Proof:
By Midpoint Theorem:
• DE ∥ AC and DE = ½ AC = AF = FC
• EF ∥ AB and EF = ½ AB = AD = DB
• FD ∥ BC and FD = ½ BC = BE = EC
This creates three parallelograms: ADEF, BFED, and FDCE
In △AFD and △DEF:
• AF = DE (proved above)
• FD = FD (common)
• AD = EF (proved above)
∴ △AFD ≅ △DEF (SSS)
Similarly: △AFD ≅ △FBE ≅ △EDC
Therefore, all four triangles are congruent to each other.
Q19.
3 Marks
Prove that the line segments joining the midpoints of the opposite sides of a quadrilateral bisect each other.
✓ Proof Required
Solution:
Let ABCD be a quadrilateral. Let P, Q, R, S be the midpoints of AB, BC, CD, DA respectively.
Join PR and QS. Let them intersect at O.
To Prove: O bisects both PR and QS
Proof:
Join AC (diagonal)
In △ABC:
P and Q are midpoints of AB and BC
By Midpoint Theorem: PQ ∥ AC and PQ = ½ AC
In △ADC:
S and R are midpoints of AD and DC
By Midpoint Theorem: SR ∥ AC and SR = ½ AC
∴ PQ ∥ SR and PQ = SR
∴ PQRS is a parallelogram (one pair of opposite sides equal and parallel)
Since diagonals of a parallelogram bisect each other (Theorem 8.6):
PR and QS bisect each other at O.
Let ABCD be a quadrilateral. Let P, Q, R, S be the midpoints of AB, BC, CD, DA respectively.
Join PR and QS. Let them intersect at O.
To Prove: O bisects both PR and QS
Proof:
Join AC (diagonal)
In △ABC:
P and Q are midpoints of AB and BC
By Midpoint Theorem: PQ ∥ AC and PQ = ½ AC
In △ADC:
S and R are midpoints of AD and DC
By Midpoint Theorem: SR ∥ AC and SR = ½ AC
∴ PQ ∥ SR and PQ = SR
∴ PQRS is a parallelogram (one pair of opposite sides equal and parallel)
Since diagonals of a parallelogram bisect each other (Theorem 8.6):
PR and QS bisect each other at O.
Q20.
3 Marks
ABC is a triangle right angled at C. A line through the midpoint M of hypotenuse AB and parallel to BC intersects AC at D. Show that:
(i) D is the midpoint of AC
(ii) MD ⊥ AC
(iii) CM = MA = ½ AB
(i) D is the midpoint of AC
(ii) MD ⊥ AC
(iii) CM = MA = ½ AB
✓ Proof Required
Solution:
Given: △ABC with ∠C = 90°, M is midpoint of AB, DM ∥ BC
(i) To prove: D is midpoint of AC
In △ABC:
M is the midpoint of AB (given)
DM ∥ BC (given)
By converse of Midpoint Theorem:
D must be the midpoint of AC
(ii) To prove: MD ⊥ AC
Since DM ∥ BC and ∠ACB = 90°
∴ ∠ADM = ∠ACB = 90° (corresponding angles)
∴ MD ⊥ AC
(iii) To prove: CM = MA = ½ AB
We know: MA = MB = ½ AB (M is midpoint)
In △AMD and △CMD:
• AD = CD (D is midpoint of AC)
• MD = MD (common)
• ∠ADM = ∠CDM = 90° (proved above)
∴ △AMD ≅ △CMD (SAS)
∴ AM = CM (CPCT)
Therefore: CM = MA = ½ AB
Given: △ABC with ∠C = 90°, M is midpoint of AB, DM ∥ BC
(i) To prove: D is midpoint of AC
In △ABC:
M is the midpoint of AB (given)
DM ∥ BC (given)
By converse of Midpoint Theorem:
D must be the midpoint of AC
(ii) To prove: MD ⊥ AC
Since DM ∥ BC and ∠ACB = 90°
∴ ∠ADM = ∠ACB = 90° (corresponding angles)
∴ MD ⊥ AC
(iii) To prove: CM = MA = ½ AB
We know: MA = MB = ½ AB (M is midpoint)
In △AMD and △CMD:
• AD = CD (D is midpoint of AC)
• MD = MD (common)
• ∠ADM = ∠CDM = 90° (proved above)
∴ △AMD ≅ △CMD (SAS)
∴ AM = CM (CPCT)
Therefore: CM = MA = ½ AB
Q21.
3 Marks
In parallelogram ABCD, E and F are the midpoints of sides AB and CD respectively. Show that the line segments AF and EC trisect the diagonal BD.
✓ Proof Required
Solution:
Given: ABCD is a parallelogram, E and F are midpoints of AB and CD
Let AF and EC intersect BD at P and Q respectively
To Prove: BP = PQ = QD
Proof:
Since ABCD is a parallelogram:
AB ∥ CD and AB = CD
E and F are midpoints, so:
AE = ½ AB and CF = ½ CD
∴ AE = CF
Also, AE ∥ CF (as AB ∥ CD)
∴ AECF is a parallelogram
In △ABP, E is the midpoint of AB and EQ ∥ AP
∴ Q is the midpoint of BP (by Midpoint Theorem converse)
∴ BQ = QP
Similarly, in △DQC:
F is midpoint of DC and FP ∥ QC
∴ P is midpoint of QD
∴ QP = PD
Therefore: BP = PQ = QD
Hence, BD is trisected.
Given: ABCD is a parallelogram, E and F are midpoints of AB and CD
Let AF and EC intersect BD at P and Q respectively
To Prove: BP = PQ = QD
Proof:
Since ABCD is a parallelogram:
AB ∥ CD and AB = CD
E and F are midpoints, so:
AE = ½ AB and CF = ½ CD
∴ AE = CF
Also, AE ∥ CF (as AB ∥ CD)
∴ AECF is a parallelogram
In △ABP, E is the midpoint of AB and EQ ∥ AP
∴ Q is the midpoint of BP (by Midpoint Theorem converse)
∴ BQ = QP
Similarly, in △DQC:
F is midpoint of DC and FP ∥ QC
∴ P is midpoint of QD
∴ QP = PD
Therefore: BP = PQ = QD
Hence, BD is trisected.
Q22.
3 Marks
Show that the bisectors of angles of a parallelogram form a rectangle.
✓ Proof Required
Solution:
Let ABCD be a parallelogram. Let the bisectors of angles A, B, C, D intersect to form quadrilateral PQRS.
To Prove: PQRS is a rectangle
Proof:
In parallelogram ABCD:
∠A + ∠B = 180° (adjacent angles are supplementary)
The bisectors divide these angles:
∠SAB = ½∠A and ∠ABP = ½∠B
In △ABS:
∠SAB + ∠ABP + ∠ASB = 180°
½∠A + ½∠B + ∠ASB = 180°
½(∠A + ∠B) + ∠ASB = 180°
½(180°) + ∠ASB = 180°
90° + ∠ASB = 180°
∠ASB = 90°
But ∠ASB = ∠P (at point P where bisectors meet)
∴ ∠P = 90°
Similarly, we can prove ∠Q = ∠R = ∠S = 90°
Therefore, PQRS is a rectangle (all angles = 90°).
Let ABCD be a parallelogram. Let the bisectors of angles A, B, C, D intersect to form quadrilateral PQRS.
To Prove: PQRS is a rectangle
Proof:
In parallelogram ABCD:
∠A + ∠B = 180° (adjacent angles are supplementary)
The bisectors divide these angles:
∠SAB = ½∠A and ∠ABP = ½∠B
In △ABS:
∠SAB + ∠ABP + ∠ASB = 180°
½∠A + ½∠B + ∠ASB = 180°
½(∠A + ∠B) + ∠ASB = 180°
½(180°) + ∠ASB = 180°
90° + ∠ASB = 180°
∠ASB = 90°
But ∠ASB = ∠P (at point P where bisectors meet)
∴ ∠P = 90°
Similarly, we can prove ∠Q = ∠R = ∠S = 90°
Therefore, PQRS is a rectangle (all angles = 90°).
SECTION D - Long Answer Questions II (5 marks each)
Q23.
5 Marks
ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD respectively. Show that:
(i) △APB ≅ △CQD
(ii) AP = CQ
(iii) Prove that APCQ is a parallelogram
(i) △APB ≅ △CQD
(ii) AP = CQ
(iii) Prove that APCQ is a parallelogram
✓ Proof Required
Solution:
Given: ABCD is a parallelogram, AP ⊥ BD, CQ ⊥ BD
(i) To prove: △APB ≅ △CQD
In △APB and △CQD:
• ∠APB = ∠CQD = 90° (given)
• AB = CD (opposite sides of parallelogram)
• ∠ABP = ∠CDQ (alternate angles, AB ∥ CD)
∴ △APB ≅ △CQD (AAS congruence)
(ii) To prove: AP = CQ
Since △APB ≅ △CQD (proved above)
∴ AP = CQ (CPCT)
(iii) To prove: APCQ is a parallelogram
We have proved: AP = CQ
Also, AP ⊥ BD and CQ ⊥ BD
∴ AP ∥ CQ (both perpendicular to same line)
Since one pair of opposite sides is equal and parallel:
APCQ is a parallelogram (by theorem)
Given: ABCD is a parallelogram, AP ⊥ BD, CQ ⊥ BD
(i) To prove: △APB ≅ △CQD
In △APB and △CQD:
• ∠APB = ∠CQD = 90° (given)
• AB = CD (opposite sides of parallelogram)
• ∠ABP = ∠CDQ (alternate angles, AB ∥ CD)
∴ △APB ≅ △CQD (AAS congruence)
(ii) To prove: AP = CQ
Since △APB ≅ △CQD (proved above)
∴ AP = CQ (CPCT)
(iii) To prove: APCQ is a parallelogram
We have proved: AP = CQ
Also, AP ⊥ BD and CQ ⊥ BD
∴ AP ∥ CQ (both perpendicular to same line)
Since one pair of opposite sides is equal and parallel:
APCQ is a parallelogram (by theorem)
Q24.
5 Marks
In △ABC, E is the midpoint of median AD. Show that area of △BED = ¼ area of △ABC.
✓ Proof Required
Solution:
Given: In △ABC, AD is a median, E is midpoint of AD
To Prove: ar(△BED) = ¼ ar(△ABC)
Proof:
Step 1: Since AD is a median of △ABC:
D is the midpoint of BC
∴ ar(△ABD) = ½ ar(△ABC) ... (1)
(Median divides triangle into two equal areas)
Step 2: In △ABD, E is the midpoint of AD
Join BE
BE is a median of △ABD (from B to midpoint of AD)
∴ ar(△BED) = ½ ar(△ABD) ... (2)
Step 3: From equations (1) and (2):
ar(△BED) = ½ ar(△ABD)
ar(△BED) = ½ × [½ ar(△ABC)]
ar(△BED) = ¼ ar(△ABC)
Hence proved.
Given: In △ABC, AD is a median, E is midpoint of AD
To Prove: ar(△BED) = ¼ ar(△ABC)
Proof:
Step 1: Since AD is a median of △ABC:
D is the midpoint of BC
∴ ar(△ABD) = ½ ar(△ABC) ... (1)
(Median divides triangle into two equal areas)
Step 2: In △ABD, E is the midpoint of AD
Join BE
BE is a median of △ABD (from B to midpoint of AD)
∴ ar(△BED) = ½ ar(△ABD) ... (2)
Step 3: From equations (1) and (2):
ar(△BED) = ½ ar(△ABD)
ar(△BED) = ½ × [½ ar(△ABC)]
ar(△BED) = ¼ ar(△ABC)
Hence proved.
Q25.
5 Marks
Show that the line segments joining the midpoints of the sides of a triangle form four triangles, each of which is congruent to the original triangle and has an area equal to one-fourth of the area of the original triangle.
✓ Proof Required
Solution:
Given: △ABC with D, E, F as midpoints of BC, CA, AB respectively
To Prove:
(i) △AFE ≅ △FBD ≅ △EDC ≅ △DEF
(ii) ar(each small triangle) = ¼ ar(△ABC)
Part (i): Proving Congruence
By Midpoint Theorem:
• EF ∥ BC and EF = ½ BC = BD = DC
• FD ∥ AC and FD = ½ AC = AE = EC
• DE ∥ AB and DE = ½ AB = AF = FB
In △AFE, △FBD, △EDC, and △DEF:
• AF = FB = BD = DC = EF
• AE = EC = FD = DE
• FE = FE (common)
By SSS congruence:
△AFE ≅ △FBD ≅ △EDC ≅ △DEF
Part (ii): Proving Area
Since EF ∥ BC:
△AEF and △ABC have same height from A
ar(△AEF)/ar(△ABC) = EF/BC
ar(△AEF)/ar(△ABC) = (½BC)/BC = ½
But this is incorrect reasoning. Let's use medians:
Since D, E, F are midpoints:
• AD is median of △ABC, so ar(△ABD) = ½ ar(△ABC)
• FD is median of △ABD, so ar(△FBD) = ½ ar(△ABD)
• Therefore: ar(△FBD) = ½ × ½ ar(△ABC) = ¼ ar(△ABC)
Since all four triangles are congruent:
Each has area = ¼ ar(△ABC)
Verification: 4 × ¼ = 1 (entire triangle)
Given: △ABC with D, E, F as midpoints of BC, CA, AB respectively
To Prove:
(i) △AFE ≅ △FBD ≅ △EDC ≅ △DEF
(ii) ar(each small triangle) = ¼ ar(△ABC)
Part (i): Proving Congruence
By Midpoint Theorem:
• EF ∥ BC and EF = ½ BC = BD = DC
• FD ∥ AC and FD = ½ AC = AE = EC
• DE ∥ AB and DE = ½ AB = AF = FB
In △AFE, △FBD, △EDC, and △DEF:
• AF = FB = BD = DC = EF
• AE = EC = FD = DE
• FE = FE (common)
By SSS congruence:
△AFE ≅ △FBD ≅ △EDC ≅ △DEF
Part (ii): Proving Area
Since EF ∥ BC:
△AEF and △ABC have same height from A
ar(△AEF)/ar(△ABC) = EF/BC
ar(△AEF)/ar(△ABC) = (½BC)/BC = ½
But this is incorrect reasoning. Let's use medians:
Since D, E, F are midpoints:
• AD is median of △ABC, so ar(△ABD) = ½ ar(△ABC)
• FD is median of △ABD, so ar(△FBD) = ½ ar(△ABD)
• Therefore: ar(△FBD) = ½ × ½ ar(△ABC) = ¼ ar(△ABC)
Since all four triangles are congruent:
Each has area = ¼ ar(△ABC)
Verification: 4 × ¼ = 1 (entire triangle)
Q26.
5 Marks
ABCD is a quadrilateral in which P, Q, R and S are midpoints of the sides AB, BC, CD and DA respectively. AC is a diagonal. Show that:
(i) SR ∥ AC and SR = ½ AC
(ii) PQ = SR
(iii) PQRS is a parallelogram
(i) SR ∥ AC and SR = ½ AC
(ii) PQ = SR
(iii) PQRS is a parallelogram
✓ Proof Required
Solution:
Given: ABCD is a quadrilateral with P, Q, R, S as midpoints of AB, BC, CD, DA
(i) To prove: SR ∥ AC and SR = ½ AC
In △ADC:
S is the midpoint of AD (given)
R is the midpoint of DC (given)
By Midpoint Theorem (Theorem 8.8):
SR ∥ AC and SR = ½ AC ... (1)
(ii) To prove: PQ = SR
In △ABC:
P is the midpoint of AB (given)
Q is the midpoint of BC (given)
By Midpoint Theorem:
PQ ∥ AC and PQ = ½ AC ... (2)
From (1) and (2):
SR = ½ AC and PQ = ½ AC
∴ PQ = SR ... (3)
(iii) To prove: PQRS is a parallelogram
From (1) and (2):
SR ∥ AC and PQ ∥ AC
∴ SR ∥ PQ ... (4)
From (3) and (4):
PQ = SR and PQ ∥ SR
Since one pair of opposite sides is equal and parallel:
PQRS is a parallelogram (by Theorem 8.3)
Note: This proves that the quadrilateral formed by joining the midpoints of any quadrilateral is always a parallelogram.
Given: ABCD is a quadrilateral with P, Q, R, S as midpoints of AB, BC, CD, DA
(i) To prove: SR ∥ AC and SR = ½ AC
In △ADC:
S is the midpoint of AD (given)
R is the midpoint of DC (given)
By Midpoint Theorem (Theorem 8.8):
SR ∥ AC and SR = ½ AC ... (1)
(ii) To prove: PQ = SR
In △ABC:
P is the midpoint of AB (given)
Q is the midpoint of BC (given)
By Midpoint Theorem:
PQ ∥ AC and PQ = ½ AC ... (2)
From (1) and (2):
SR = ½ AC and PQ = ½ AC
∴ PQ = SR ... (3)
(iii) To prove: PQRS is a parallelogram
From (1) and (2):
SR ∥ AC and PQ ∥ AC
∴ SR ∥ PQ ... (4)
From (3) and (4):
PQ = SR and PQ ∥ SR
Since one pair of opposite sides is equal and parallel:
PQRS is a parallelogram (by Theorem 8.3)
Note: This proves that the quadrilateral formed by joining the midpoints of any quadrilateral is always a parallelogram.
Quick Answer Reference - Section A
Q1: (b) Parallelogram
Q2: (a) 70°
Q3: (b) Equal and bisect each other
Q4: (b) Rhombus
Q5: (b) Half of the third side and parallel to it
Q6: (b) Two congruent triangles
