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Congruent triangles and Relationship with triangles
Chapter 5 & 6 — Congruent Triangles & Relationships within Triangles
Clear class notes, step-by-step proofs, worked examples and practice questions for Big Ideas Math — Common Core Geometry (Level 12).
Overview
These notes cover Chapter 5 — Congruent Triangles and Chapter 6 — Relationships within Triangles. Focus: formal congruence proofs (SSS, SAS, ASA, AAS, RHS), properties derived from congruence (medians, perpendicular bisectors, angle bisectors), and geometric reasoning using congruent parts of triangles.
Estimated class time: 2–3 lessons. Use the worked examples to practise proof structure and the practice questions to test fluency.
Key Theorems & Postulates
SSS (Side-Side-Side)
If three sides of one triangle are equal to three sides of another, the triangles are congruent.
SAS (Side-Angle-Side)
If two sides and the included angle of one triangle equal two sides and included angle of another, the triangles are congruent.
ASA (Angle-Side-Angle)
If two angles and the included side of one triangle equal two angles and the included side of another, they are congruent.
AAS (Angle-Angle-Side)
If two angles and a non-included side of one triangle equal the corresponding parts of another triangle, they are congruent.
RHS (Right angle-Hypotenuse-Side)
In right triangles: if hypotenuse and one leg are equal, triangles are congruent.
Consequences of Congruence
Corresponding parts are equal (CPCTC): corresponding sides and angles match. Used to prove medians, perpendicular bisectors, angle bisectors, and altitudes relationships.
Worked Examples (step-by-step)
Example 1 — RHS on an isosceles base
Given: In triangle ABC, AB = AC and AD ⟂ BC. Prove that triangle ABD ≅ triangle ACD and that D is the midpoint of BC.
Proof structure:
- Note that AD is common to triangles ABD and ACD.
- AB = AC (given).
- Angles ADB and ADC are right angles (AD ⟂ BC).
- So each triangle has a right angle, equal hypotenuse (AB = AC), and an equal leg (AD common). By RHS, triangles are congruent.
- CPCTC gives BD = DC, so D is the midpoint of BC.
Example 2 — Diagonals in a rhombus
Given: ABCD is a rhombus. Diagonals AC and BD intersect at O. Prove that triangles AOB, BOC, COD and DOA are congruent.
Proof structure:
- In a rhombus, diagonals bisect each other: AO = OC and BO = OD.
- All sides equal: AB = BC = CD = DA. In particular AB = BC.
- Consider triangles AOB and BOC: AO = OC (diagonal bisection), BO common, AB = BC (sides of rhombus). By SSS, triangles AOB ≅ BOC. Repeat for others.
- Hence all four are congruent; angles and side relations follow.
Practice Questions (Congruency only)
Try these on paper. Solutions are given at the end of the page.
- In triangle ABC, AB = AC and AD ⟂ BC. Prove △ABD ≅ △ACD and show D is midpoint of BC.
- Triangles PQR and XYZ have PQ = XY, QR = YZ, PR = XZ. Prove congruence and name the postulate.
- In triangle ABC, point D on AB and E on AC satisfy AD = AE and ∠ADB = ∠AEC. Prove triangles ADB and AEC are congruent.
- The diagonals of a rhombus intersect at O. Show the 4 triangles formed are congruent (state postulate).
- In triangle PQR, PQ = PR and ∠Q = ∠R. A line from Q meets PR at S such that ∠PQS = ∠PSR. Prove △PQS ≅ △PSR.
Teaching Tips & Common Mistakes
- Always label the triangles you will compare; write corresponding vertices in the same order (e.g., △ABC vs △DEF).
- Don't confuse similarity with congruence — similar triangles have proportional sides, congruent triangles have equal sides/angles.
- Be explicit: name the postulate you use (SSS, SAS, ASA, AAS, RHS) and write CPCTC when using corresponding parts.
- When using angle bisector or perpendicular bisector facts, state the property clearly ("perpendicular bisector = equidistant from endpoints").
Solutions (Congruency Problems)
Note: These solutions give one clear proof approach. There can be valid alternative proofs — check for logical equivalence.
Solution 1
Triangles ABD and ACD: AB = AC (given), AD common, ADB and ADC are right angles. By RHS, triangles are congruent. So BD = DC.
Solution 2
Three corresponding sides equal → SSS → triangles congruent.
Solution 3
Use AAS: AD = AE, ∠ADB = ∠AEC, and angle at A common (or can be shown equal by construction). So triangles ADB and AEC are congruent.
Solution 4
Diagonals bisect each other and sides equal → SSS (or SAS) for adjacent triangles → all four congruent.
Solution 5
Using given equal sides and equal angles construct triangles PQS and PSR: show two sides and included angle equal (SAS) or use other matching facts to get congruence; hence QS = SR.
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Geometry Test — Congruent Triangles & Relationships within Triangles
$1This comprehensive geometry test features two sections — Congruent Triangles and Relationships within Triangles. It challenges students with advanced, multi-step reasoning and coordinate geometry problems while reinforcing understanding of triangle centers, medians, bisectors, and geometric properties.
Section A – Congruent Triangles
Total Marks: 35
- (6 marks) In △ABC, AB = AC and AD ⟂ BC. Prove that △ABD ≅ △ACD and show that D is the midpoint of BC.
- (6 marks) In △PQR, PQ = PR and PS bisects ∠QPR. If QS = RS, prove that △PQS ≅ △PRS and hence show that PS ⟂ QR.
- (8 marks) The diagonals of a rhombus ABCD intersect at O. Prove that △AOB ≅ △BOC ≅ △COD ≅ △DOA. Hence, deduce two properties of a rhombus.
- (6 marks) In △ABC, D and E are midpoints of sides AB and AC respectively. Prove that DE ∥ BC and DE = ½BC. Explain why △ADE ≅ △ABC is not true even though they are similar.
- (9 marks) In △XYZ, medians from X, Y, Z intersect at G. Given coordinates X(0, 0), Y(6, 0), and Z(2, 4), find coordinates of G and show that medians divide each other in the ratio 2:1.
Section B – Relationships within Triangles
Total Marks: 35
- (6 marks) In △ABC, perpendicular bisectors of sides AB and AC meet at O. Prove that OA = OB = OC. Hence, state the name of point O.
- (6 marks) In △ABC, the angle bisectors of ∠A, ∠B, and ∠C meet at I. Prove that I is equidistant from all sides of the triangle.
- (8 marks) Triangle PQR has coordinates P(0, 0), Q(8, 0), and R(4, 6). Find the equations of medians and show that they are concurrent. Determine the coordinates of centroid G.
- (8 marks) In △ABC, sides AB and AC are extended to points D and E such that AD = AE. Prove that △ADE ≅ △CAB. Hence, show that DE ∥ BC.
- (7 marks) The perpendicular bisectors of triangle PQR meet at O(2, 3). Given Q(6, 7) and R(0, 7), find coordinates of P and prove that PQ = QR = RP.
Solutions
Section A Solutions – Congruent Triangles
Q1: Use RHS Congruence: AB = AC, AD common, ∠ADB = ∠ADC = 90°. Thus △ABD ≅ △ACD ⇒ BD = DC.
Q2: AB = AC, BD = CD, AD common → SSS. Hence △ABD ≅ △ACD and AD ⟂ BC.
Q3: Diagonals bisect and sides equal → SSS ⇒ all 4 triangles congruent. Properties: diagonals bisect and are perpendicular.
Q4: DE ∥ BC and DE = ½BC (Midpoint Theorem). Triangles are similar, not congruent, since sides not equal.
Q5: Median intersection point G = ( (x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3 ) = (8/3, 4/3).
Section B Solutions – Relationships within Triangles
Q6: Perpendicular bisectors meet at circumcenter; OA = OB = OC proven by RHS congruence.
Q7: Angle bisectors intersect at incenter I. Distances from I to sides equal ⇒ I equidistant from all sides.
Q8: Use midpoint formula and find median equations; intersection = centroid, dividing each in 2:1.
Q9: Equal sides AD = AE and ∠DAE = ∠CAB → SAS → triangles congruent. Hence DE ∥ BC.
Q10: Using perpendicular bisector equations, find intersection satisfying equal distance from all vertices ⇒ equilateral triangle formed.