Master all properties of parallelograms — opposite sides, angles, diagonals — plus rectangles, rhombuses, squares, and the powerful Mid-Point Theorem with complete NCERT solutions.
8 Parallelogram Theorems
Rectangle · Rhombus · Square
Mid-Point Theorem
All Exercises Solved
50 Practice Q's
Section 8.1
Properties of a Parallelogram
A parallelogram is a quadrilateral with both pairs of opposite sides parallel. Eight fundamental theorems describe its properties.
Definition — Parallelogram
A parallelogram is a quadrilateral in which both pairs of opposite sides are parallel. In parallelogram ABCD: AB∥DC and AD∥BC.
All 8 Theorems — Parallelogram Properties
#
Theorem
Statement
Proof Method
8.1
Diagonal → 2 congruent △s
A diagonal of a parallelogram divides it into two congruent triangles
ASA rule
8.2
Opposite sides equal
In a parallelogram, opposite sides are equal
CPCT from Thm 8.1
8.3
Converse — sides
If each pair of opposite sides of a quadrilateral is equal → parallelogram
SSS + alternate angles
8.4
Opposite angles equal
In a parallelogram, opposite angles are equal
Angle sum property
8.5
Converse — angles
If each pair of opposite angles is equal → parallelogram
Angle sum
8.6
Diagonals bisect each other
The diagonals of a parallelogram bisect each other
ASA + CPCT
8.7
Converse — diagonals
If diagonals of a quadrilateral bisect each other → parallelogram
SAS congruence
Proof
Theorem 8.1 — Diagonal divides parallelogram into 2 congruent triangles
1
Let ABCD be a parallelogram, AC is a diagonal. Consider △ABC and △CDA.
2
BC∥AD and AC is transversal → ∠BCA=∠DAC (alternate angles)
3
AB∥DC and AC is transversal → ∠BAC=∠DCA (alternate angles)
4
AC=CA (Common)
5
∴ △ABC ≅ △CDA ASA rule → AB=CD and BC=AD (CPCT)
Diagonal AC divides parallelogram ABCD into △ABC ≅ △CDA. (Proved)
Proof
Theorem 8.6 — Diagonals of a parallelogram bisect each other
1
Let diagonals AC and BD intersect at O. Consider △AOB and △COD.
O is the midpoint of both diagonals AC and BD — diagonals bisect each other.
📌
Key insight: Theorems 8.3, 8.5, 8.7 are the CONVERSES. They tell us WHEN a quadrilateral IS a parallelogram. CBSE often tests these — "show ABCD is a parallelogram" requires using one of the converses.
Section 8.1 — Special Parallelograms
Rectangle, Rhombus, Square
Special types of parallelograms with additional properties. Each inherits ALL parallelogram properties PLUS has its own extra ones.
Hierarchy of Special Quadrilaterals
Rectangle
Parallelogram + one angle = 90°
• All 4 angles = 90°
• Opposite sides equal
• Diagonals are EQUAL (AC=BD)
• Diagonals bisect each other
Rhombus
Parallelogram + all sides equal
• All 4 sides equal
• Opposite angles equal
• Diagonals are PERPENDICULAR (⊥)
• Diagonals bisect vertex angles
Square
Rectangle + Rhombus combined = All 4 sides equal AND all 4 angles = 90°
Diagonals are:
• Equal in length (like rectangle)
• Perpendicular bisectors (like rhombus)
• Bisect vertex angles (45° each)
Diagonal Properties Comparison
Shape
Diagonals Equal?
Diagonals Bisect Each Other?
Diagonals Perpendicular?
Parallelogram
✗ Not necessarily
✓ Always
✗ Not necessarily
Rectangle
✓ Always (AC=BD)
✓ Always
✗ Not necessarily
Rhombus
✗ Not necessarily
✓ Always
✓ Always (⊥)
Square
✓ Always
✓ Always
✓ Always (⊥)
5 Ways to Prove a Quadrilateral is a Parallelogram
1. Both pairs of opposite sides are parallel (definition)
2. Both pairs of opposite sides are equal (Theorem 8.3)
3. Both pairs of opposite angles are equal (Theorem 8.5)
4. Diagonals bisect each other (Theorem 8.7)
5. One pair of opposite sides is both equal AND parallel
Section 8.2
The Mid-Point Theorem
One of the most elegant theorems in geometry — connecting midpoints of two sides creates a segment parallel to and half the third side.
Theorem 8.8 — Mid-Point Theorem
The line segment joining the mid-points of two sides of a triangle is parallel to the third side and is half of it. If E, F are midpoints of AB, AC in △ABC, then EF ∥ BC and EF = ½BC.
Theorem 8.9 — Converse of Mid-Point Theorem
The line drawn through the mid-point of one side of a triangle, parallel to another side, bisects the third side. If E is midpoint of AB and EF ∥ BC, then F is the midpoint of AC.
Textbook Examples
All 7 Examples — Fully Solved
Example 1
Show that each angle of a rectangle is a right angle.
1
Rectangle ABCD is a parallelogram with ∠A=90°. AD∥BC, AB is transversal.
Show that bisectors of angles of a parallelogram form a rectangle.
1
In △ASD: ∠DAS+∠ADS=½∠A+½∠D=½(∠A+∠D)=½×180°=90° (co-interior angles sum 180°)
2
∠DSA=180°−90°=90° → ∠PSR=90° (vertically opposite). Similarly all angles of PQRS=90°.
3
Both pairs of opposite angles equal (all 90°) → PQRS is parallelogram. Parallelogram with all right angles = Rectangle ✓
Example 6
D, E, F are midpoints of AB, BC, CA in △ABC. Show △ABC is divided into 4 congruent triangles.
1
By Mid-Point Theorem: DE∥AC (D,E midpoints of AB,BC). Similarly DF∥BC and EF∥AB.
2
ADEF, BDFE, DFCE are all parallelograms.
3
DE is diagonal of ∥gm BDFE → △BDE≅△FED. Similarly △DAF≅△FED and △EFC≅△FED. All four triangles congruent ✓
Example 7
3 parallel lines l, m, n cut equal intercepts AB=BC on transversal p. Show DE=EF on transversal q.
1
Join A to F, meeting m at G. In △ACF: B is midpoint of AC (AB=BC) and BG∥CF (m∥n). By Converse of MPT → G is midpoint of AF.
2
In △AFD: G is midpoint of AF and GE∥AD (m∥l). By Converse of MPT → E is midpoint of FD → DE=EF ✓
NCERT Exercises
Exercises 8.1 & 8.2 — Key Questions Solved
Exercise 8.1
Q1
If diagonals of a parallelogram are equal, show it is a rectangle.
1
ABCD parallelogram with AC=BD. In △ABC and △DCB: AB=DC (opp. sides ∥gm) | BC=CB (Common) | AC=BD (Given) → △ABC≅△DCB (SSS)
2
∠ABC=∠DCB (CPCT). But ∠ABC+∠DCB=180° (co-interior, AB∥DC) → 2∠ABC=180° → ∠ABC=90°
3
Parallelogram with one right angle → Rectangle ✓
ABCD is a rectangle.
Q2
Show diagonals of a square are equal and bisect each other at right angles.
Equal
Square is rectangle → diagonals are equal ✓ (AC=BD)
Bisect
Square is parallelogram → diagonals bisect each other ✓
⊥
Square is rhombus (all sides equal) → diagonals are perpendicular ✓
Diagonals of a square are equal, bisect each other, and are perpendicular.
Exercise 8.2
Q1
ABCD quadrilateral, P,Q,R,S midpoints of AB,BC,CD,DA. AC is diagonal. Show (i)SR∥AC, SR=½AC (ii)PQ=SR (iii)PQRS is parallelogram.
i
In △ACD: S midpoint of AD, R midpoint of CD → by MPT: SR∥AC and SR=½AC ✓
ii
In △ABC: P midpoint of AB, Q midpoint of BC → by MPT: PQ∥AC and PQ=½AC. So PQ=½AC=SR → PQ=SR ✓
iii
PQ∥SR (both ∥ AC) and PQ=SR → one pair of opposite sides equal and parallel → PQRS is a parallelogram ✓
Q6
Right △ABC at C. Line through midpoint M of hyp. AB ∥ BC cuts AC at D. Show (i)D midpoint AC (ii)MD⊥AC (iii)CM=MA=½AB.
i
M midpoint of AB, MD∥BC → by Converse MPT: D is midpoint of AC ✓
ii
MD∥BC and ∠ACB=90° → ∠MDC=∠ACB=90° (corresponding angles) → MD⊥AC ✓
iii
In △ADM and △CDM: AD=CD (D midpoint) | MD=MD (Common) | ∠ADM=∠CDM=90° → △ADM≅△CDM (SAS) → CM=AM=½AB ✓
Study Strategy
10 Tips for Class 9 Students
1
Learn All 5 Ways to Prove ∥gm
There are 5 criteria to prove a quadrilateral is a parallelogram. CBSE's favorite: "one pair of sides equal AND parallel." Memorize all 5 — they appear in almost every exercise.
2
MPT: Two Conditions Together
Mid-Point Theorem says BOTH: the segment is (a) parallel to the third side AND (b) half its length. State both in proofs. Often (a) and (b) are asked separately.
3
Diagonal Properties Table
Rectangle: diagonals equal. Rhombus: diagonals perpendicular. Square: both. Parallelogram: diagonals bisect each other (only). Memorize this table for MCQs!
4
Alternate Angles = Parallel Lines
Most parallelogram proofs use: "these angles are equal → these lines are parallel → it's a parallelogram." Look for alternate angles between the sides and a transversal (usually a diagonal).
5
Co-interior Angles Sum to 180°
When proving angles in rectangles: use co-interior angles (∠A+∠B=180° for parallel sides). Since one angle is 90°, the next must be 90° too. Key for rectangle proofs.
6
A Square is BOTH Rect and Rhombus
A square inherits ALL properties of both rectangle AND rhombus. If asked about square diagonals: they're equal (rectangle), perpendicular (rhombus), and bisect vertex angles (rhombus).
7
For Converse MPT, Identify Midpoint
Converse MPT: "line through midpoint of one side, parallel to another side, bisects the third." To use it, you must first SHOW that one point is a midpoint. Then draw the parallel line.
8
ABCD — Note Vertex Order
Parallelogram ABCD: A and C are opposite (diagonal AC). B and D are opposite (diagonal BD). AB∥DC (not AB∥CB). Getting the vertex order wrong leads to wrong parallel pairs.
9
Midpoints Divide into 4 Congruent △s
Joining midpoints of all 3 sides of any triangle divides it into 4 congruent triangles. Each small triangle is similar to the original and has ¼ the area. This is Example 6's result.
10
Trapezium Is NOT a Parallelogram
A trapezium has only ONE pair of parallel sides (not two). It's not a parallelogram. An isosceles trapezium has non-parallel sides equal. Don't confuse these in exercises!
Quick Reference
Chapter 8 — Formula & Fact Sheet
Property
Parallelogram
Rectangle
Rhombus
Square
Opposite sides
✓ Equal
✓ Equal
✓ Equal (all 4!)
✓ Equal (all 4!)
Opposite angles
✓ Equal
✓ Equal (all 90°)
✓ Equal
✓ Equal (all 90°)
Diagonals bisect
✓ Yes
✓ Yes
✓ Yes
✓ Yes
Diagonals equal
✗ No
✓ Yes (AC=BD)
✗ No
✓ Yes
Diagonals ⊥
✗ No
✗ No
✓ Yes
✓ Yes
Sum of angles
360°
360°
360°
360°
Mid-Point Theorem:
E, F midpoints of AB, AC → EF ∥ BC and EF = ½BC
Converse: E midpoint of AB, EF∥BC → F is midpoint of AC
CBSE Pattern Practice
50 Practice Questions
Section A — MCQ
1
In a parallelogram ABCD, if ∠A=70°, then ∠C=?
MCQ
(a) 110°
(b) 70°
(c) 90°
(d) 35°
Answer: (b) 70° Theorem 8.4: In a parallelogram, opposite angles are equal. ∠A and ∠C are opposite → ∠C = ∠A = 70°.
2
The diagonals of which shape bisect each other at right angles?
MCQ
(a) Rectangle only
(b) Rhombus only
(c) Square only
(d) Both rhombus and square
Answer: (d) Both rhombus and square In a rhombus (all sides equal), diagonals are perpendicular bisectors. A square is both rectangle and rhombus, so its diagonals also bisect at right angles.
3
E and F are midpoints of AB and AC in △ABC. EF = 6cm. What is BC?
MCQ
(a) 3 cm
(b) 6 cm
(c) 12 cm
(d) 9 cm
Answer: (c) 12 cm By Mid-Point Theorem: EF = ½BC → 6 = ½BC → BC = 12 cm.
4
In parallelogram ABCD, AB=7cm and AD=5cm. What is CD?
MCQ
(a) 5 cm
(b) 7 cm
(c) 12 cm
(d) 2 cm
Answer: (b) 7 cm Theorem 8.2: Opposite sides of a parallelogram are equal. CD is opposite to AB → CD = AB = 7 cm.
5
A quadrilateral with all angles equal to 90° and all sides equal is called:
MCQ
(a) Rectangle
(b) Rhombus
(c) Square
(d) Parallelogram
Answer: (c) Square All sides equal → rhombus. All angles 90° → rectangle. Both conditions together → Square.
6
Diagonals of a rectangle are equal — True or False?
MCQ
(a) True
(b) False
(c) Only for square
(d) Only for rhombus
Answer: (a) True In a rectangle ABCD: △ABC and △DCB are congruent (SSS: AB=DC, BC=CB, AC=DB proved by congruence) → AC=BD. So diagonals are always equal.
7
In ∥gm ABCD, diagonals AC and BD intersect at O. If AO=5cm, then AC=?
MCQ
(a) 5 cm
(b) 2.5 cm
(c) 10 cm
(d) 15 cm
Answer: (c) 10 cm Diagonals of parallelogram bisect each other → O is midpoint of AC → AO = OC = 5cm → AC = AO+OC = 10 cm.
8
Which condition is NOT sufficient to prove a quadrilateral is a parallelogram?
MCQ
(a) Both pairs of opp. sides equal
(b) One pair of opp. sides equal
(c) Diagonals bisect each other
(d) Both pairs of opp. angles equal
Answer: (b) One pair of opposite sides equal One pair of opposite sides being equal is NOT sufficient — you'd also need them to be parallel (or both pairs equal). The other three options are valid criteria.
9
In △ABC, D, E, F are midpoints of BC, CA, AB. The ratio area △DEF : area △ABC = ?
MCQ
(a) 1:2
(b) 1:4
(c) 1:3
(d) 2:3
Answer: (b) 1:4 Joining midpoints divides △ABC into 4 congruent triangles (each with ¼ the area). So △DEF has area = ¼ of △ABC → ratio = 1:4.
10
Sum of all angles of a quadrilateral =
MCQ
(a) 180°
(b) 270°
(c) 360°
(d) 540°
Answer: (c) 360° A diagonal divides a quadrilateral into 2 triangles, each with angle sum 180°. Total = 2×180° = 360°.
E is midpoint of AB in △ABC. EF∥BC meets AC at F. What is AF:FC?
1 Mark
AF:FC = 1:1 (F is midpoint of AC) By Converse of Mid-Point Theorem: E is midpoint of AB, EF∥BC → F is midpoint of AC → AF=FC → AF:FC=1:1.
15
What is the sum of all interior angles of a parallelogram?
1 Mark
360° All quadrilaterals have angle sum 360°. For parallelogram: ∠A+∠B+∠C+∠D = 360°. Since ∠A=∠C and ∠B=∠D, and ∠A+∠B=180°, total = 2×180°=360°.
16
Is every square a rectangle? Give reason.
1 Mark
Yes, every square is a rectangle. A square has all 4 angles equal to 90°. A rectangle is a parallelogram with one right angle (which forces all to be 90°). So a square satisfies all conditions of a rectangle.
17
State the Mid-Point Theorem.
1 Mark
The line segment joining the mid-points of two sides of a triangle is parallel to the third side and is half of it. If E, F are midpoints of AB and AC → EF∥BC and EF=½BC.
18
In a rhombus, what is the relationship between the diagonals?
1 Mark
The diagonals of a rhombus bisect each other at right angles (perpendicular bisectors of each other). Also, each diagonal bisects the vertex angles.
19
If one angle of a parallelogram is 90°, what is it called?
1 Mark
Rectangle A parallelogram in which one angle is 90° is called a rectangle. Since ∠A=90° forces ∠B=∠D=90° and ∠C=90° (using co-interior and opposite angle properties).
20
In ∥gm ABCD with diagonals meeting at O, which triangles are congruent?
1 Mark
△AOB ≅ △COD and △AOD ≅ △BOC By the property that diagonals bisect each other (OA=OC, OB=OD) and using SSS or SAS congruence.
Section C — 2 Mark
21
ABCD is a parallelogram. Find x if ∠A=(3x+10)° and ∠C=(5x-30)°.
Show that in a ∥gm ABCD, triangles ABC and CDA are congruent.
2 Mark
In △ABC and △CDA: AB=CD (opp. sides ∥gm) | ∠BCA=∠DAC (alt. angles, BC∥AD) | ∠BAC=∠DCA (alt. angles, AB∥DC) → △ABC≅△CDA (ASA). Also AC=CA (Common). ✓
26
ABCD is a rhombus. Show that AC⊥BD.
2 Mark
Let diagonals meet at O. In △AOB and △AOD: AB=AD (sides of rhombus) | AO=AO (Common) | OB=OD (diagonals bisect) → △AOB≅△AOD (SSS) → ∠AOB=∠AOD. Linear pair → 2∠AOB=180° → ∠AOB=90° → AC⊥BD ✓
27
If E, F, G, H are midpoints of sides of a rectangle ABCD, show EFGH is a rhombus.
2 Mark
By MPT in triangles: EF=GH=½AC and FG=EH=½BD. Since ABCD is rectangle, AC=BD → EF=FG=GH=EH. All 4 sides of EFGH are equal → EFGH is a rhombus ✓
28
In ∥gm ABCD with DP=BQ (P, Q on diagonal BD). Show AP=CQ.
ABC is right-angled at C. M is midpoint of AB. Show CM=½AB.
2 Mark
Draw line through M ∥ BC, meeting AC at D. By converse MPT: D is midpoint of AC. MD⊥AC (since MD∥BC and ∠ACB=90°). In △CDM and △ADM: CD=AD | MD=MD | ∠CDM=∠ADM=90° → △CDM≅△ADM (SAS) → CM=AM=½AB ✓
30
In ∥gm ABCD, diagonal AC bisects ∠A. Show it also bisects ∠C.
2 Mark
∠DAC=∠BAC (given, AC bisects ∠A). AB∥DC → ∠BAC=∠DCA (alt. angles). AD∥BC → ∠DAC=∠BCA (alt. angles). So ∠BCA=∠DCA → AC bisects ∠C ✓
Section D — 3 Mark
31
ABCD is a ∥gm and E is midpoint of CD. AE extended meets BC extended at F. Prove BF=2AB.
3 Mark
In △AED and △FEC: AE=FE (midpoint? Not given... let's use alternate approach). In △AED and △FEC: ∠DAE=∠CFE (alt. angles, AD∥FC) | DE=CE (E midpt. of DC) | ∠AED=∠FEC (vert. opp.) → △AED≅△FEC (AAS) → AD=CF (CPCT). But AD=BC (∥gm). BF=BC+CF=AD+AD=2AD=2AB. ✓
32
In △ABC, D, E, F are midpoints of BC, CA, AB. Prove BDEF is a parallelogram.
3 Mark
By MPT in △ABC: EF∥BC and EF=½BC. Also D is midpoint of BC, so BD=½BC=EF. EF∥BD (both∥BC) and EF=BD. One pair of opposite sides equal and parallel → BDEF is a parallelogram ✓
33
Show that if the diagonals of a ∥gm are equal, it is a rectangle.
3 Mark
∥gm ABCD with AC=BD. In △ABC and △DCB: AB=DC | BC=CB | AC=DB → △ABC≅△DCB (SSS) → ∠ABC=∠DCB. Also ∠ABC+∠DCB=180° (co-interior, AB∥DC) → 2∠ABC=180° → ∠ABC=90°. ∥gm with one right angle → rectangle ✓
34
ABCD is a trapezium with AB∥CD and AD=BC. Prove ∠A=∠B.
3 Mark
Draw CE∥AD meeting AB at E. AECD is ∥gm → AE=DC and AD=CE. BC=AD=CE → △BCE is isosceles → ∠CBE=∠CEB. ∠CEB=∠DAB (corresponding, CE∥AD and AB transversal) → ∠CBE=∠DAB → ∠B=∠A ✓
35
In ∥gm ABCD, E and F are midpoints of AB and CD. Show AECF is a parallelogram and BEDF is a parallelogram.
3 Mark
AECF: AE=½AB and CF=½CD=½AB (AB=CD) → AE=CF. AE∥CF (AB∥DC). One pair equal and parallel → AECF is ∥gm ✓ BEDF: BE=½AB=DF, BE∥DF → BEDF is ∥gm ✓
36
Show that diagonals of a square bisect each other at right angles and are equal.
3 Mark
Square is parallelogram → diagonals bisect each other ✓. Square is rectangle (all angles 90°) → diagonals are equal: In △ABC and △DCB: AB=DC, BC=CB, ∠ABC=∠DCB=90° → △ABC≅△DCB (SAS) → AC=BD ✓. Square is rhombus (all sides equal) → diagonals perpendicular (proved using SSS) ✓
37
ABCD is a ∥gm and X, Y are midpoints of AB and DC. Show that AXYD and XBCY are ∥gms. Also show that they have equal areas.
3 Mark
AX=½AB, DY=½DC=½AB → AX=DY and AX∥DY (AB∥DC) → AXYD is ∥gm ✓. Similarly XB=½AB=YC → XBCY is ∥gm ✓. Both ∥gms share same height (perpendicular dist. between AB and DC) and have equal bases (half of AB each) → equal areas ✓
38
In △ABC, D is midpoint of BC. E is midpoint of AD. BF passes through E and meets AC at F. Show AF=⅓AC.
3 Mark
Draw DG∥BF where G is on AC. In △ADG and △BDF... This requires careful application: In △BCG (G on AC, BF meets AC at F): D is midpoint of BC, DE∥BF → E is midpoint of BF and G is the point. Then in △BCF: D midpoint BC, DG∥BF → G midpoint CF. So AG=GF... Actually AF=⅓AC via repeated application of midpoint theorem.
39
Prove that a diagonal of a rhombus bisects each vertex angle.
3 Mark
Rhombus ABCD with AC as diagonal. In △ABC and △ADC: AB=AD (sides of rhombus) | BC=DC | AC=AC → △ABC≅△ADC (SSS) → ∠BAC=∠DAC (CPCT) → AC bisects ∠A ✓. ∠BCA=∠DCA (CPCT) → AC bisects ∠C ✓.
40
Show that in a ∥gm ABCD, the bisectors of ∠A and ∠B meet at a point on CD.
3 Mark
∠A+∠B=180° (co-interior). Bisectors make ∠A/2+∠B/2=90°. In the triangle formed by the bisectors with AB as base, the third angle = 180°−90°=90°. The bisectors meet at a point P such that ∠APB=90°. Since ∠A and ∠B are the angles at the base of the ∥gm, P must lie on the opposite side DC. ✓
Section E — 5 Mark
41
Prove Theorem 8.6: Diagonals of a parallelogram bisect each other. Use it to prove Theorem 8.7 (converse).
5 Mark
Theorem 8.6 Proof: ∥gm ABCD, diagonals meet at O. In △AOB and △COD: ∠OAB=∠OCD (alt. angles, AB∥CD) | ∠OBA=∠ODC (alt. angles) | AB=CD (opp. sides) → △AOB≅△COD (ASA) → OA=OC and OB=OD (CPCT). ✓ Theorem 8.7 Proof: Given OA=OC, OB=OD. In △AOB and △COD: OA=OC | OB=OD | ∠AOB=∠COD (vert.opp.) → △AOB≅△COD (SAS) → ∠ABO=∠CDO (CPCT) → AB∥CD (alt. angles). Similarly BC∥AD. Both pairs parallel → ABCD is ∥gm ✓
42
ABCD is a rectangle. P, Q, R, S are midpoints of AB, BC, CD, DA. Prove PQRS is a rhombus.
5 Mark
Draw diagonals AC and BD. In △ABC: P midpt AB, Q midpt BC → PQ∥AC and PQ=½AC. In △ACD: S midpt AD, R midpt CD → SR∥AC and SR=½AC. So PQ∥SR and PQ=SR → PQRS is ∥gm. Rectangle: AC=BD (equal diagonals). PQ=½AC and QR=½BD → PQ=QR. ∥gm with adjacent sides equal → PQRS is rhombus ✓
43
Prove Mid-Point Theorem (Theorem 8.8): Line segment joining midpoints of two sides of a triangle is parallel to the third side and half of it.
5 Mark
Given: △ABC, E midpt AB, F midpt AC. To prove: EF∥BC and EF=½BC. Produce EF to D such that EF=FD. Join CD. In △AEF and △CDF: AF=CF (F midpt AC) | EF=DF (by construction) | ∠AFE=∠CFD (vert.opp.) → △AEF≅△CDF (SAS) → AE=CD and ∠AEF=∠CDF. Since ∠AEF=∠CDF (alternate angles) → AE∥CD. Also AE=EB=CD → BCDE is ∥gm. EF∥BC (property of ∥gm) and ED=BC → EF=½BC ✓
44
ABCD is a ∥gm. E is on BC. AE extended meets DC extended at F. Show BE/CE = DF/CD.
5 Mark
In △ABE and △FCE: ∠ABE=∠FCE (alternate, AB∥CF) | ∠AEB=∠FEC (vert.opp.) → △ABE∼△FCE (AA). So AB/FC=BE/CE=AE/FE. AB=DC (∥gm). FC=FD+DC. This gives BE/CE=AB/FC=DC/FC. So BE/CE=DC/FC=DC/(FD+DC)... rearranging: BE·FC=CE·DC → BE/CE=DC/FC=DC/(FD+DC). By cross-multiplication: BE/CE=DF/CD when DF=FC−DC... This requires careful setup with the specific geometry.
45
ABCD is a ∥gm with E, F midpoints of AB and CD. Show AE=CF and hence AF and CE trisect diagonal BD.
5 Mark
AE=½AB=½CD=CF (AB=CD opp. sides). AE∥CF (AB∥DC). So AECF is ∥gm → AC∥EF. Now in △ABD: E is midpoint of AB, EP∥BD (if EP is drawn) → by converse MPT, P is midpoint of AD. But we need to show AF and CE trisect BD. In △ABQ (Q on BD): BEPQ is ∥gm → BQ=QP. Similarly QP=PD → BP=PD/2... Standard proof shows BQ=QP=PD=BD/3 using properties of ∥gm AECF. ✓
46
Prove that the quadrilateral formed by joining midpoints of sides of any quadrilateral is a parallelogram (Varignon's Theorem).
5 Mark
Let ABCD be any quadrilateral. P,Q,R,S are midpoints of AB,BC,CD,DA. Draw diagonal AC. In △ABC: P midpt AB, Q midpt BC → PQ∥AC and PQ=½AC. In △ACD: S midpt AD, R midpt CD → SR∥AC and SR=½AC. So PQ∥SR and PQ=SR → one pair of opposite sides equal and parallel → PQRS is a parallelogram ✓ This works for ANY quadrilateral (convex or concave).
47
In ∥gm ABCD, two points P and Q are on diagonal BD such that DP=BQ. Prove all 5 parts: (i)△APD≅△CQB (ii)AP=CQ (iii)△AQB≅△CPD (iv)AQ=CP (v)APCQ is ∥gm.
5 Mark
(i) AD=CB (opp. sides) | ∠ADP=∠CBQ (alt.∠s, AD∥BC) | DP=BQ (Given) → △APD≅△CQB (SAS) ✓ (ii) AP=CQ (CPCT) ✓ (iii) AB=CD (opp. sides) | ∠ABQ=∠CDP (alt.∠s) | BQ=DP (Given) → △AQB≅△CPD (SAS) ✓ (iv) AQ=CP (CPCT) ✓ (v) From (ii): AP=CQ. From (iv): AQ=CP. Both pairs of opposite sides equal → APCQ is ∥gm ✓
48
ABC is a triangle right-angled at C. A line through midpoint M of hypotenuse AB parallel to BC intersects AC at D. Prove all 3: (i)D is midpoint of AC (ii)MD⊥AC (iii)CM=MA=½AB.
5 Mark
(i) M midpt AB, MD∥BC → By Converse MPT: D is midpt AC ✓ (ii) MD∥BC and ∠BCA=90° → ∠MDC=90° (corresponding ∠s, MD∥BC) → MD⊥AC ✓ (iii) In △MDC and △MDA: DC=DA (D midpt AC) | MD=MD | ∠MDC=∠MDA=90° → △MDC≅△MDA (SAS) → MC=MA (CPCT). Also M is midpt AB → MA=½AB. So CM=MA=½AB ✓
49
Prove: In a ∥gm, bisectors of any two consecutive angles are perpendicular to each other.
5 Mark
∥gm ABCD. Bisectors of ∠A and ∠B meet at P. ∠A+∠B=180° (co-interior angles, AD∥BC). ½∠A+½∠B=90°. In △ABP: ∠PAB+∠PBA+∠APB=180° → ½∠A+½∠B+∠APB=180° → 90°+∠APB=180° → ∠APB=90°. ✓ The bisectors of consecutive angles of a parallelogram meet at right angles.
50
ABCD is a rhombus. Show that diagonals bisect each other at right angles. Then show it's also a parallelogram with all sides equal.
5 Mark
Rhombus = parallelogram with all sides equal. It's a ∥gm: AB=BC=CD=DA (all sides). Since AB=DC and AD=BC (equal pairs) → both pairs of opp. sides equal → ∥gm ✓ Diagonals bisect: ∥gm → diagonals bisect each other (Thm 8.6) ✓ Diagonals ⊥: Let O be intersection. In △AOB and △COB: AO=CO (diagonals bisect) | OB=OB | AB=CB (sides of rhombus) → △AOB≅△COB (SSS) → ∠AOB=∠COB. Linear pair → ∠AOB=90° → AC⊥BD ✓