Master congruence of triangles — all 5 criteria (SAS, ASA, AAS, SSS, RHS), isosceles triangle theorems, and all NCERT exercises fully solved with step-by-step proofs.
SAS · ASA · AAS
SSS · RHS
Isosceles Theorems
CPCT
50 Practice Q's
Section 7.2
Congruence of Triangles
Two figures are congruent if they are exactly the same shape and size — one can be placed over the other to cover it completely.
Definition — Congruent Figures
Two figures are congruent if they have the same shape and the same size (equal in all respects). When △PQR ≅ △ABC, the correspondence is P↔A, Q↔B, R↔C and the corresponding sides and angles are equal.
CPCT — Corresponding Parts of Congruent Triangles
When two triangles are proved congruent, ALL their corresponding parts (sides AND angles) are automatically equal. We use the shorthand CPCT to state this. This is one of the most powerful tools in geometry proofs.
⚠️
Order matters! △PQR ≅ △ABC means P↔A, Q↔B, R↔C. Writing △QRP ≅ △ABC would be WRONG for the same triangles. Always write vertices in the correct corresponding order.
Section 7.3
Congruence Criteria — SAS, ASA, AAS
These are the rules that let us prove two triangles congruent. Each requires knowing specific matching parts.
Axiom 7.1 — SAS (Side-Angle-Side)
Two triangles are congruent if two sides and the included angle of one triangle are equal to the two sides and the included angle of the other triangle. Note: The angle MUST be between (included by) the two sides. SSA (angle not included) does NOT work!
Theorem 7.1 — ASA (Angle-Side-Angle)
Two triangles are congruent if two angles and the included side of one triangle are equal to the two angles and the included side of the other triangle. (Proved using SAS axiom)
AAS (Angle-Angle-Side)
Two triangles are congruent if two angles and one side (not necessarily included) of one triangle are equal to the corresponding two angles and side of the other triangle. Why? If two angle pairs are equal, the third pair is also equal (sum = 180°), so it reduces to ASA.
SAS vs ASS/SSA — Why Order Matters
In SAS, the angle must be the included angle (between the two equal sides). If the angle is NOT between the equal sides (SSA or ASS), the triangles may NOT be congruent. This is a common exam mistake!
Rule
What's Equal
Works?
Notes
SAS
2 sides + included angle
✓ YES
Angle must be between the 2 sides
ASA
2 angles + included side
✓ YES
Side must be between the 2 angles
AAS
2 angles + any corresponding side
✓ YES
Side need not be included
SSA/ASS
2 sides + non-included angle
✗ NO
Triangles may differ!
AAA
3 angles
✗ NO
Only similar, not necessarily congruent
Section 7.4
Properties of Isosceles Triangles
An isosceles triangle has exactly two equal sides. Its angle properties are fundamental and appear in many proofs.
Definition — Isosceles Triangle
A triangle with two equal sides is called an isosceles triangle. The two equal sides are called the legs, and the third side is the base. The angles at the base (opposite to equal sides) are called base angles.
Theorem 7.2 — Isosceles Triangle Theorem
Angles opposite to equal sides of an isosceles triangle are equal. If AB = AC in △ABC, then ∠B = ∠C.
Proof
Theorem 7.2 — Using Angle Bisector
1
Given: △ABC with AB = AC. Draw the bisector of ∠A; let it meet BC at D.
2
In △BAD and △CAD: AB = AC (Given) | ∠BAD = ∠CAD (By construction, AD bisects ∠A) | AD = AD (Common)
3
Therefore △BAD ≅ △CAD SAS rule
4
So ∠ABD = ∠ACD (CPCT) → ∠B = ∠C ✓
∠B = ∠C (Proved by SAS congruence)
Theorem 7.3 — Converse of Isosceles Triangle Theorem
Sides opposite to equal angles of a triangle are equal. If ∠B = ∠C in △ABC, then AB = AC. (Proved using ASA congruence)
💡
Equilateral triangle corollary: Each angle of an equilateral triangle is 60°. (Since all 3 sides are equal, all angles are equal, and they sum to 180° → each = 60°.)
Section 7.5
SSS and RHS Congruence
Two more powerful congruence criteria, completing the full set of 5 rules.
Theorem 7.4 — SSS (Side-Side-Side)
If three sides of one triangle are equal to the three sides of another triangle, then the two triangles are congruent.
Theorem 7.5 — RHS (Right angle-Hypotenuse-Side)
If in two right triangles, the hypotenuse and one side of one triangle are equal to the hypotenuse and one side of the other triangle, then the two triangles are congruent. Note: RHS is a special case. The right angle is NOT the included angle — yet it works!
Complete Summary of All 5 Congruence Criteria
SAS: 2 sides + included ∠ASA: 2 ∠s + included sideAAS: 2 ∠s + any sideSSS: 3 sidesRHS: hypotenuse + side (right △)
Textbook Examples
All Examples — Fully Solved
All 8 NCERT examples from Chapter 7, with complete step-by-step proofs.
Example 1
OA=OB, OD=OC. Show (i)△AOD≅△BOC (ii)AD∥BC
i
In △AOD and △BOC: OA=OB (Given) | OD=OC (Given) | ∠AOD=∠BOC (Vertically opposite angles) ∴ △AOD ≅ △BOC SAS rule
ii
∠OAD = ∠OBC (CPCT). These are alternate angles for AD and BC with transversal AB. ∴ AD ∥ BC
△AOD ≅ △BOC (SAS); AD ∥ BC (alternate angles)
Example 2
AB is a line segment, l is its perpendicular bisector. P lies on l. Show PA=PB.
1
Line l⊥AB through midpoint C of AB. In △PCA and △PCB: AC=BC (C is midpoint) | ∠PCA=∠PCB=90° (Given) | PC=PC (Common)
2
∴ △PCA ≅ △PCB SAS rule → PA=PB (CPCT)
PA = PB — any point on the perpendicular bisector is equidistant from both endpoints.
Example 3
AB∥CD, O is midpoint of AD. Show (i)△AOB≅△DOC (ii)O is midpoint of BC.
Show that angles of an equilateral triangle are 60° each.
1
In equilateral △ABC: AB=BC=CA. Since AB=AC → ∠B=∠C (Thm 7.2). Since AB=BC → ∠A=∠C.
2
So ∠A=∠B=∠C. Sum = 180° → 3∠A=180° → ∠A=60°.
Each angle of an equilateral triangle = 60°.
Study Strategy
10 Tips for Class 9 Students
1
Write Vertex Order Correctly
△ABC ≅ △PQR means A↔P, B↔Q, C↔R. The ORDER of vertices tells you which parts correspond. Always match the correct vertices — wrong order = wrong congruence statement.
2
State the Rule Every Time
After stating the three equal parts, always write "By SAS/ASA/AAS/SSS/RHS rule". CBSE gives separate marks for citing the correct congruence rule. Never omit it.
3
SAS: Angle MUST be Included
In SAS, the angle must be BETWEEN (sandwiched by) the two equal sides. SSA/ASS doesn't work. If someone gives you two sides and a non-included angle, you cannot conclude congruence.
4
CPCT — Use it After Proof
You can only use CPCT AFTER proving the triangles are congruent. Never use CPCT before establishing congruence — that would be circular reasoning.
5
Common Side = "Common"
When a side is shared by both triangles (like AC in △ABC and △ACD), state it as "AC = AC (Common)". Don't leave it unstated — CBSE needs all 3 conditions listed.
6
AAA ≠ Congruent
Three equal angles only means the triangles are SIMILAR (same shape), not congruent (same size). For congruence, at least one pair of sides must be equal.
7
Isosceles: Equal Sides → Equal Angles
Always identify the isosceles triangle first. If AB=AC, then ∠B=∠C (Theorem 7.2). Conversely if ∠B=∠C, then AB=AC (Theorem 7.3). Both directions are used frequently.
8
Draw Clear Diagrams
For every proof question, draw a large, clear, labelled diagram. Mark tick marks for equal sides and arcs for equal angles. CBSE awards marks for correct diagrams.
9
RHS Only for Right Triangles
RHS (Right angle-Hypotenuse-Side) works only when there is a 90° angle. The hypotenuse is the side OPPOSITE the right angle (longest side). A regular SSA with a non-90° angle doesn't work.
10
Vertically Opposite Angles
When two lines cross, vertically opposite angles are equal. This is often the key "equal angle" that establishes SAS or AAS congruence. Always look for line intersections in diagrams.
Quick Reference
Chapter 7 — Formula & Fact Sheet
Concept
Statement
Rule Used
SAS
2 sides + included ∠ equal
Axiom 7.1
ASA
2 ∠s + included side equal
Theorem 7.1
AAS
2 ∠s + any corresponding side
Derived from ASA
SSS
3 sides equal
Theorem 7.4
RHS
Hypotenuse + one side (right △)
Theorem 7.5
Isosceles (→)
AB=AC → ∠B=∠C
Theorem 7.2
Isosceles (←)
∠B=∠C → AB=AC
Theorem 7.3
Equilateral
Each angle = 60°
Corollary of Thm 7.2
CPCT
Use AFTER proving congruence
Definition
Perp. bisector
Any pt on ⊥ bisector is equidistant from endpoints
SAS + CPCT
CBSE Pattern Practice
50 Practice Questions
Section A — MCQ (1 mark each)
1
Which congruence rule is an axiom (not provable from others)?
MCQ
(a) ASA
(b) SAS
(c) SSS
(d) AAS
Answer: (b) SAS SAS is taken as an axiom. ASA, AAS, SSS, RHS are theorems proved using SAS.
2
In △ABC, AB = AC. Then:
MCQ
(a) ∠A = ∠B
(b) ∠B = ∠C
(c) ∠A = ∠C
(d) ∠A = ∠B = ∠C
Answer: (b) ∠B = ∠C Theorem 7.2: Angles opposite to equal sides (AB and AC) are equal. The angles opposite AB is ∠C, and opposite AC is ∠B. So ∠B = ∠C.
3
△ABC ≅ △DEF under correspondence A↔D, B↔E, C↔F. Which is correct?
Answer: (c) SSA SSA (two sides and a non-included angle) is NOT a valid congruence criterion. The angle must be included between the sides (SAS) for it to work.
5
In right △ABC, right-angled at B, hypotenuse AC = 5 cm and BC = 3 cm. AB = ?
MCQ
(a) 3 cm
(b) 4 cm
(c) 5 cm
(d) 2 cm
Answer: (b) 4 cm By Pythagoras: AB² + BC² = AC² → AB² + 9 = 25 → AB² = 16 → AB = 4 cm.
6
CPCT stands for:
MCQ
(a) Congruent Parts of Congruent Triangles
(b) Corresponding Parts of Congruent Triangles
(c) Common Parts of Congruent Triangles
(d) None of these
Answer: (b) Corresponding Parts of Congruent Triangles CPCT = Corresponding Parts of Congruent Triangles. It means all corresponding sides and angles of two congruent triangles are equal.
7
Each angle of an equilateral triangle is:
MCQ
(a) 45°
(b) 60°
(c) 90°
(d) 120°
Answer: (b) 60° In equilateral △ABC: AB=BC=CA → all angles equal. Sum = 180° → each = 60°.
8
RHS congruence rule applies only to:
MCQ
(a) Acute triangles
(b) Obtuse triangles
(c) Right triangles
(d) All triangles
Answer: (c) Right triangles RHS (Right angle-Hypotenuse-Side) specifically requires a 90° angle. It only applies to right-angled triangles.
9
If ∠B = ∠C = 70° in △ABC, which sides are equal?
MCQ
(a) AB = BC
(b) BC = CA
(c) AB = CA
(d) None
Answer: (c) AB = CA Theorem 7.3: Sides opposite to equal angles are equal. Side opp ∠B is AC (= CA). Side opp ∠C is AB. Since ∠B = ∠C, AB = CA.
10
Three equal angles are given for two triangles. Are they congruent?
MCQ
(a) Yes always
(b) Only if one side is also equal
(c) Never
(d) Only equilateral triangles
Answer: (b) Only if one side is also equal AAA only proves similarity, not congruence. For congruence (AAS or ASA), at least one pair of corresponding sides must also be equal.
Section B — 1 Mark
11
In △PQR ≅ △XYZ, which angle corresponds to ∠Q?
1 Mark
∠Y In △PQR ≅ △XYZ, the correspondence is P↔X, Q↔Y, R↔Z. So ∠Q corresponds to ∠Y.
12
State Theorem 7.2 (Isosceles Triangle Theorem).
1 Mark
Angles opposite to equal sides of an isosceles triangle are equal. If AB = AC in △ABC, then ∠ABC = ∠ACB (i.e., ∠B = ∠C).
13
Is SSA a valid congruence criterion? Give reason.
1 Mark
No, SSA is NOT valid. When two sides and a non-included angle are equal, two different triangles can be formed. The angle must be included between the two equal sides (SAS) for congruence to be guaranteed.
14
In isosceles △ABC with AB=AC, if ∠A=100°, find ∠B.
RHS = Right angle – Hypotenuse – Side In two right triangles, if the hypotenuse and one side of one triangle equal the hypotenuse and one side of the other, the triangles are congruent.
16
If △ABC ≅ △PQR and AB = 5cm, find PQ.
1 Mark
PQ = 5 cm In △ABC ≅ △PQR, A↔P, B↔Q, C↔R. So AB corresponds to PQ. CPCT → PQ = AB = 5 cm.
17
Can three equal angles (AAA) prove congruence of triangles?
1 Mark
No. AAA only proves that triangles are similar (same shape), not congruent (same size). Two triangles can have the same angles but different sizes.
18
Name the congruence criterion used when two angles and the side between them are equal.
1 Mark
ASA (Angle-Side-Angle) Two triangles are congruent if two angles and the included side of one equal two angles and the included side of the other.
19
In right △ABC at B, which side is the hypotenuse?
1 Mark
AC is the hypotenuse. The hypotenuse is the side opposite the right angle. Right angle is at B, so the side opposite B is AC.
20
If ∠B = ∠C in △ABC, which two sides are equal?
1 Mark
AB = AC Theorem 7.3 (converse of Thm 7.2): Sides opposite to equal angles are equal. Side opposite ∠B is AC, side opposite ∠C is AB. Since ∠B=∠C, AB=AC.
Section C — 2 Mark
21
In △ABC, ∠B=45°, ∠C=45°. Show AB=AC.
2 Mark
∠B = ∠C = 45°. By Theorem 7.3 (sides opposite equal angles are equal): Side opposite ∠B is AC. Side opposite ∠C is AB. Since ∠B = ∠C → AB = AC. ∴ △ABC is isosceles with AB = AC.
22
In △ABC ≅ △DEF, find the value of x if BC = (2x+3) cm and EF = (x+7) cm.
2 Mark
Since △ABC ≅ △DEF, BC = EF (CPCT) 2x+3 = x+7 → x = 4. BC = EF = 11 cm.
23
In △PQR, PQ=PR. The bisector of ∠P meets QR at M. Show QM=MR.
2 Mark
In △PQM and △PRM: PQ=PR (Given) | ∠QPM=∠RPM (PM bisects ∠P) | PM=PM (Common) ∴ △PQM ≅ △PRM (SAS) → QM=MR (CPCT). M is the midpoint of QR.
24
Two triangles ABC and DEF have AB=DE=5cm, BC=EF=7cm and CA=FD=6cm. Which congruence rule applies?
2 Mark
SSS (Side-Side-Side) Congruence Rule All three sides of △ABC equal the corresponding three sides of △DEF: AB=DE=5, BC=EF=7, CA=FD=6. By Theorem 7.4 (SSS rule): △ABC ≅ △DEF.
25
In △ABC, right-angled at C, if AC=BC, find ∠A and ∠B.
In △ABC with ∠A=40°, if it is isosceles with AB=BC, find the other two angles.
2 Mark
AB=BC → angles opp. to them are equal → ∠C=∠A=40° (∠A is opp. BC, ∠C is opp. AB). ∠B=180°−40°−40°=∠B=100°, ∠C=40°.
27
OA=OB. OC and OD are bisectors of ∠AOC and ∠BOC resp. passing through O. Show OC=OD.
2 Mark
By symmetry: In △OAC and △OBD (with OC⊥OA and OD⊥OB): OA=OB (Given), ∠OCA=∠ODB=90°, OC=OD (if the angles are bisected equally). Full proof uses RHS: OA=OB, right angles equal, AC=BD by angle bisector property.
28
In △ABC, if ∠B=30° and ∠C=30°, what type of triangle is it? Find ∠A.
2 Mark
Isosceles triangle (∠B=∠C → AB=AC by Thm 7.3). ∠A = 180°−30°−30° = 120°. It is an obtuse isosceles triangle.
29
D is the midpoint of BC in △ABC. Show △ABD ≅ △ACD if AB=AC.
2 Mark
In △ABD and △ACD: AB=AC (Given) | BD=CD (D is midpoint of BC) | AD=AD (Common) ∴ △ABD ≅ △ACD (SSS rule)
30
In right △ABC at B and right △DEF at E, AC=DF=5cm and BC=EF=3cm. Show △ABC≅△DEF.
2 Mark
In △ABC and △DEF: ∠B=∠E=90° (Right angles) | AC=DF=5cm (Hypotenuses) | BC=EF=3cm (Sides) ∴ △ABC ≅ △DEF (RHS rule)
Section D — 3 Mark
31
In △ABC, AB=AC. D is a point on BC such that AD⊥BC. Prove AD bisects ∠A.
3 Mark
In △ADB and △ADC: AB=AC (Given) | AD=AD (Common) | ∠ADB=∠ADC=90° (AD⊥BC) ∴ △ADB ≅ △ADC (RHS rule) → ∠BAD=∠CAD (CPCT) → AD bisects ∠A.
32
In quadrilateral ABCD, AB=CD and AB∥CD. Prove AC=BD.
3 Mark
Draw AC and BD. In △ABC and △DCB: AB=DC (Given) | ∠ABC=∠DCB (AB∥DC, alternate interior angles... wait, need AB∥CD with transversal BC) | BC=CB (Common). ∴ △ABC ≅ △DCB (SAS) → AC=DB=BD (CPCT). ✓
33
Prove: In △ABC with AB=AC, if D is midpoint of BC, then AD⊥BC.
Two sides AB and BC and median AM of △ABC are respectively equal to sides PQ and QR and median PN of △PQR. Prove △ABC≅△PQR.
3 Mark
M is midpoint of BC → BM=½BC. N is midpoint of QR → QN=½QR. Since BC=QR → BM=QN. In △ABM and △PQN: AB=PQ (Given) | BM=QN (Shown) | AM=PN (Given) → △ABM≅△PQN (SSS) → ∠B=∠Q (CPCT). Now in △ABC and △PQR: AB=PQ, ∠B=∠Q, BC=QR → △ABC≅△PQR (SAS). ✓
35
ABCD is a quadrilateral, AB=AD and CB=CD. Prove AC bisects ∠A and ∠C.
3 Mark
In △ABC and △ADC: AB=AD (Given) | CB=CD (Given) | AC=AC (Common) → △ABC≅△ADC (SSS). ∠BAC=∠DAC (CPCT) → AC bisects ∠A ✓ ∠BCA=∠DCA (CPCT) → AC bisects ∠C ✓
36
In △ABC, AB=AC. Point D is on AB and E is on AC such that AD=AE. Prove △BEC≅△CDB.
3 Mark
AB=AC and AD=AE → AB−AD=AC−AE → DB=EC. In △BEC and △CDB: BE=CD (Since AB=AC, AD=AE → AB−AD=AC−AE → DB=EC, and BE=AC−AE, CD=AB−AD) | BC=CB (Common) | ∠B=∠C (Angles opp. equal sides AB,AC). ∴ △BEC≅△CDB (SAS). ✓
37
Show that in a right triangle, the hypotenuse is the longest side.
3 Mark
In right △ABC with ∠B=90°: ∠A+∠C=90° (Since ∠A+∠B+∠C=180°) → each of ∠A and ∠C < 90° = ∠B. In a triangle, the greater angle has the longer opposite side: ∠B > ∠A → AC > BC. And ∠B > ∠C → AC > AB. So AC (hypotenuse, opposite the right angle) > both legs. ✓
38
In △ABC with ∠A=60°, AB=AC. Prove △ABC is equilateral.
3 Mark
AB=AC → ∠B=∠C (Thm 7.2, angles opp. equal sides). ∠A+∠B+∠C=180° → 60°+∠B+∠B=180° → 2∠B=120° → ∠B=∠C=60°. All angles = 60° → all sides equal → △ABC is equilateral. ✓
39
P is the midpoint of AB in △ABC. Q is the midpoint of AC. PQ is drawn. If △ABC is isosceles with AB=AC, show △APQ is also isosceles.
3 Mark
P midpoint of AB → AP = AB/2. Q midpoint of AC → AQ = AC/2. Since AB=AC → AB/2=AC/2 → AP=AQ. In △APQ: AP=AQ → △APQ is isosceles with AP=AQ. Also ∠APQ=∠AQP. ✓
40
In isosceles △ABC (AB=AC), D is a point on BC. Prove AD cannot be the longest side.
3 Mark
In △ABD: ∠ABD + ∠ADB + ∠DAB = 180°. Since ∠ADB > 90° is possible... Better approach: D is ON BC, so AD < AB (since in △ABD, ∠ABD < ∠ADB when D is interior, or by the fact that a straight line is the shortest path). The median/cevian is always shorter than the lateral sides unless the triangle is degenerate. AD < AB = AC, so AD cannot be the longest side.
Section E — 5 Mark
41
Prove Theorem 7.1 (ASA congruence rule) using the SAS axiom. Consider all three cases.
5 Mark
Given: △ABC and △DEF with ∠B=∠E, ∠C=∠F, BC=EF. To prove: △ABC≅△DEF. Case I: AB=DE. Then in △ABC and △DEF: AB=DE, ∠B=∠E, BC=EF → △ABC≅△DEF by SAS. ✓ Case II: AB>DE. Take P on AB with PB=DE. In △PBC and △DEF: PB=DE, ∠B=∠E, BC=EF → △PBC≅△DEF (SAS) → ∠PCB=∠DFE. But ∠ACB=∠DFE (Given) → ∠ACB=∠PCB → P coincides with A → AB=DE → reduces to Case I. ✓ Case III: ABIn all cases △ABC≅△DEF. QED
42
Prove: In △ABC, if AB=AC, then (i) the perpendicular bisector of BC passes through A (ii) the bisector of ∠A bisects BC at right angles.
5 Mark
(i) Let D be midpoint of BC. In △ABD and △ACD: AB=AC (Given) | BD=CD (D is midpoint) | AD=AD (Common) → △ABD≅△ACD (SSS) → ∠ADB=∠ADC (CPCT). Linear pair → ∠ADB=90°. So AD⊥BC and D is midpoint → AD is the perpendicular bisector of BC. A lies on it. ✓ (ii) From above, AD bisects BC (D is midpoint) and AD⊥BC. We also have ∠BAD=∠CAD (CPCT from △ABD≅△ACD) → AD bisects ∠A. ✓ So the bisector of ∠A bisects BC at right angles.
43
In right △ABC right-angled at C, M is the midpoint of hypotenuse AB. C is joined to M and produced to D such that DM=CM. D is joined to B. Prove all 4 results of Exercise 7.1 Q8.
5 Mark
(i) △AMC≅△BMD: AM=BM (M midpoint of AB) | CM=DM (Given) | ∠AMC=∠BMD (vert. opp.) → SAS → △AMC≅△BMD ✓ (ii) ∠DBC=90°: ∠DBC=∠ACB (CPCT from i). ∠ACB=90° (Given) → ∠DBC=90° ✓ (iii) △DBC≅△ACB: DB=AC (CPCT from i) | ∠DBC=∠ACB=90° | BC=CB (Common) → SAS → △DBC≅△ACB ✓ (iv) CM=½AB: DC=AB (CPCT from iii). DM=CM → DC=DM+MC=CM+CM=2CM → AB=2CM → CM=½AB ✓
44
△ABC and △DBC are two isosceles triangles on same base BC, vertices A and D on same side. If AD extended intersects BC at P, prove: (i)△ABD≅△ACD (ii)△ABP≅△ACP (iii)AP bisects ∠A and ∠D (iv)AP⊥BC.
5 Mark
(i) △ABC isosceles: AB=AC. △DBC isosceles: DB=DC. In △ABD and △ACD: AB=AC | DB=DC | AD=AD → SSS → △ABD≅△ACD ✓ (ii) ∠BAP=∠CAP (CPCT from i). In △ABP and △ACP: AB=AC | ∠BAP=∠CAP | AP=AP → SAS → △ABP≅△ACP ✓ (iii) ∠BAP=∠CAP (CPCT from i) → AP bisects ∠A ✓. ∠BDP=∠CDP (CPCT from i in △DBP,△DCP similarly) → AP bisects ∠D ✓ (iv) ∠APB=∠APC (CPCT from ii). ∠APB+∠APC=180° → 2∠APB=180° → ∠APB=90° → AP⊥BC ✓. Also BP=CP (CPCT from ii) → P is midpoint → AP is ⊥ bisector of BC ✓
45
ABCD is a quadrilateral in which AD=BC and ∠DAB=∠CBA. Prove: (i)△ABD≅△BAC (ii)BD=AC (iii)∠ABD=∠BAC.
5 Mark
(i) In △ABD and △BAC: AD=BC (Given) | ∠DAB=∠CBA (Given) | AB=BA (Common) → SAS → △ABD≅△BAC ✓ (ii) BD=AC (CPCT from i) ✓ (iii) ∠ABD=∠BAC (CPCT from i) ✓
46
Prove that the angle bisectors from two equal vertices of an isosceles triangle are equal in length.
5 Mark
Let △ABC with AB=AC. Let BM bisect ∠B (M on AC) and CN bisect ∠C (N on AB). Since AB=AC → ∠B=∠C → ∠B/2=∠C/2 → ∠ABM=∠ACN. In △ABM and △ACN: AB=AC (Given) | ∠A=∠A (Common) | ∠ABM=∠ACN (Shown) → △ABM≅△ACN (AAS) → BM=CN (CPCT). ∴ The angle bisectors from B and C are equal. ✓
47
In quadrilateral ACBD, AC=AD, AB bisects ∠A. Show △ABC≅△ABD and hence BC=BD. Then prove that P, the midpoint of CD, lies on AB.
5 Mark
Part 1: In △ABC and △ABD: AC=AD (Given) | ∠CAB=∠DAB (AB bisects ∠A) | AB=AB (Common) → SAS → △ABC≅△ABD → BC=BD (CPCT). ✓ Part 2 (P on AB): From △ABC≅△ABD → ∠ACB=∠ADB (CPCT). In △ACP and △ADP: AC=AD | CP=DP (P midpoint of CD) | AP=AP (Common) → SSS → △ACP≅△ADP → ∠CAP=∠DAP. So P lies on AB (the bisector of ∠A). ✓
48
Prove the converse of the isosceles triangle theorem: if two angles of a triangle are equal, the sides opposite to them are equal.
5 Mark
Given: △ABC with ∠B=∠C. To prove: AB=AC. Draw AD bisecting ∠A, meeting BC at D. In △ABD and △ACD: ∠ABD=∠ACD (Given, ∠B=∠C) | ∠BAD=∠CAD (AD bisects ∠A) | AD=AD (Common) ∴ △ABD≅△ACD (AAS rule) → AB=AC (CPCT). ✓
49
Line l is the bisector of ∠A and B is any point on l. BP⊥ and BQ⊥ to the arms of ∠A. Prove △APB≅△AQB and BP=BQ.
5 Mark
In △APB and △AQB: ∠BAP = ∠BAQ (l bisects ∠A) ∠APB = ∠AQB = 90° (BP⊥AP, BQ⊥AQ) AB = AB (Common) ∴ △APB ≅ △AQB (AAS rule) ✓ → BP = BQ (CPCT) ✓ ∴ B is equidistant from the arms of ∠A.
50
In right △ABC (right-angled at C), prove that the median CM to hypotenuse AB equals half of AB. (Use the construction: produce CM to D with DM=CM, prove △DBC is right-angled.)
5 Mark
Step 1: Produce CM to D with DM=CM. In △AMC and △BMD: AM=BM (M midpoint) | CM=DM | ∠AMC=∠BMD (vert.opp.) → △AMC≅△BMD (SAS) → DB=CA and ∠DBC=∠ACB=90°. Step 2: So BD⊥BC. In △DBC and △ACB: DB=CA | BC=CB | ∠DBC=∠ACB=90° → △DBC≅△ACB (SAS) → DC=AB. Step 3: DC=DM+MC=2CM. So 2CM=AB → CM=½AB. ✓