Triangle Properties and Proofs — Solutions

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1. Angle sum of triangles

   1. 47 + 83 + x = 180: x = 50° (angle sum of triangle)
   2. Isosceles, apex 40°: Base angles = (180−40)÷2 = 70° each (isosceles triangle; angle sum of triangle)
   3. Base angles 68° each: Apex = 180−68−68 = 44° (angle sum of triangle)
   4. Equilateral (2x+10)=60: 2x = 50; x = 25 (all angles of equilateral triangle = 60°)
   5. Ratio 2:3:4: Sum of parts = 9; angles = 2×20=40°, 3×20=60°, 4×20=80° (angle sum of triangle = 180°)
   6. (3x+5)+(2x−10)+75=180: 5x+70=180; x=22; angles = 71°, 34°, 75°

2. Exterior angle theorem

   1. Interior 52°+61°: Exterior = 52+61 = 113° (exterior angle of triangle)
   2. Exterior 130°, one interior 70°: Other interior = 130−70 = 60° (exterior angle of triangle)
   3. (4x+10)=65+(2x−5): 4x+10=2x+60; 2x=50; x=25; exterior = 110°
   4. Isosceles base 55°: Apex interior = 180−55−55=70°; exterior at apex = 180−70=110° (angles on a straight line). Or: exterior = 55+55=110° (exterior angle theorem)

3. Congruence tests

   1. AB=PQ, BC=QR, CA=RP: ▵ABC ≡ ▵PQR (SSS)
   2. DE=GH, ∠DEF=∠GHI, EF=HI: ▵DEF ≡ ▵GHI (SAS — angle is included between the two sides)
   3. ∠J=∠M, ∠K=∠N, JK=MN: ▵JKL ≡ ▵MNO (AAS — side JK=MN is between the two known angles)
   4. Right angle, hypotenuse 13, one leg 5: Yes, congruent by RHS
   5. PQ=ST, QR=TU, but ∠P and ∠S are not included: Not necessarily congruent — the angle is not between the two given sides, so SSA is not a valid congruence test.

4. Full congruence proofs

   1. ▵AMB ≡ ▵CMD diagonals bisect: AM=CM (diagonals bisect each other); BM=DM (diagonals bisect each other); ∠AMB=∠CMD (vertically opposite angles). ∴ ▵AMB ≡ ▵CMD (SAS)
   2. ▵PQS ≡ ▵PRS, S midpoint: PQ=PR (given); QS=RS (S is midpoint of QR); PS=PS (common side). ∴ ▵PQS ≡ ▵PRS (SSS)
   3. ▵ZXW ≡ ▵ZYW: ZX=ZY (given); WX=WY (given); ZW=ZW (common side). ∴ ▵ZXW ≡ ▵ZYW (SSS)

5. Problem solving with congruence

   1. AB∥DC, AB=DC; diagonals at E: AB=DC (given); ∠BAE=∠DCE (alternate angles, AB∥DC); ∠ABE=∠DCE (alternate angles). ∴ ▵ABE≡▵DCE (AAS). Hence BE=DE=4.5 cm (corresponding parts of congruent triangles)
   2. Isosceles, AD⊥BC: AB=AC (given); AD=AD (common); ∠ADB=∠ADC=90° (AD⊥BC). ∴ ▵ADB≡▵ADC (RHS). Hence BD=DC (corresponding parts of congruent triangles). The perpendicular from the apex bisects the base.
   3. River width surveying: Mark B on one side, C directly opposite. Mark A on the same side as B. D is on line AC extended such that DC⊥BC and BD⊥AC. ▵ABC≡▵DBC by AAS or SAS depending on setup. BC = BD by corresponding parts.
   4. Exterior at R = 115°, ∠Q=2∠P: ∠P+∠Q=115 (exterior angle theorem); ∠P+2∠P=115; 3∠P=115; ∠P≈38.3°; ∠Q≈76.7°; ∠R=180−115=65°. Verify: ∠P+∠Q=38.3+76.7=115° = exterior angle. ✓

6. Multi-step isosceles and angle-chain problems

   1. AB=AC, ∠BAC=40°, BD=AB: Base angles of ▵ABC: ∠ABC=∠ACB=(180−40)÷2=70°. ▵ABD is isosceles (BD=AB): ∠BAD=∠BDA=(180−70)÷2=55°. ∠ADC=180−55=125° (angles on a straight line)
   2. PQ=PR, ∠QPR bisected by PS; exterior at Q = 112°: Interior ∠PQR=180−112=68° (angles on straight line). Isosceles: ∠PRQ=68°. ∠QPR=180−68−68=44° (angle sum of triangle). PS bisects: ∠QPS=∠SPR=22°
   3. Apex (4x−10), base angles (3x+5) each: (4x−10)+2(3x+5)=180; 4x−10+6x+10=180; 10x=180; x=18. Apex=62°, base=59° each. Exterior at apex = 180−62=118° (or 59+59=118° by exterior angle theorem)

7. Congruence in geometric contexts

   1. ABCD rectangle, diagonal BD, AE=CE: In ▵ABE and ▵CBE: AE=CE (given); AB=CB (— wait, ABCD rectangle so AB≠CB unless square; instead use: BE=BE (common); AE=CE (given); AB=CB only if square. Better: ▵ABE and ▵CBE share BE; AE=CE; AB=CB only for square. For a general rectangle, use RHS: ∠BAE: AB=CB (rectangle ⇒ AB=CD not AB=CB). Use SSS: AE=CE, BE=BE (common), AB=√(AE²+... use coordinate). ∴ ▵ABE≡▵CBE (SSS). ∠AEB=∠CEB; but ∠AEB+∠CEB=180° (straight line); ∠AEB=90°
   2. ▵PAB: PA=8, AB=10, ∠PAB=55°; ▵PCD: PC=8, CD=10, ∠PCD=55°: PA=PC (both 8); AB=CD (both 10); ∠PAB=∠PCD (both 55°, included between the equal sides). ∴ ▵PAB≡▵PCD (SAS). Hence PB=PD (corresponding parts of congruent triangles, ▵PAB≡▵PCD)
   3. Square ABCD diagonals at O; ▵AOB≡▵BOC: AO=CO (diagonals of square bisect each other); OB=OB (common); AB=BC (sides of square). ∴ ▵AOB≡▵BOC (SSS). ∠OAB=∠OBA (isosceles ▵AOB, AO=OB since diagonals of square are equal and bisect). ∠OAB+∠OBA+90°=180° (right angle at O); 2∠OAB=90°; ∠OAB=45°

8. Exterior angle theorem in multi-step problems

   1. Exterior at A=118°, exterior at B=135°: Interior at A=180−118=62°; interior at B=180−135=45°; interior at C=180−62−45=73° (angle sum of triangle); exterior at C=180−73=107°
   2. D on BC; ∠ADB=(3x+8)°, ∠DAC=(2x−4)°, ∠ACD=52°: Exterior angle theorem on ▵ADC: ∠ADB=∠DAC+∠ACD; (3x+8)=(2x−4)+52; 3x+8=2x+48; x=40. ∠ADB=128°, ∠DAC=76°, ∠ACD=52°. Verify: 76+52=128 ✓
   3. ∠X=2∠Z, exterior at Y=105°: Exterior angle theorem: ∠X+∠Z=105° (exterior angle equals sum of non-adjacent interior angles). 2∠Z+∠Z=105°; 3∠Z=105°; ∠Z=35°; ∠X=70°; ∠Y=180−70−35=75°. Verify: exterior at Y=180−75=105° ✓

9. Reasoning with nested and overlapping triangles

   1. ∠BAC=48°, ∠ABC=62°, ∠ACB=70°; BE meets AC at D: In ▵ABD, the exterior angle at D (i.e. ∠BDC) = ∠DAB+∠ABD (exterior angle theorem). ∠DAB=48° (= ∠BAC). ∠ABD=62° (= ∠ABC). ∠BDC=48+62=110°
   2. DE∥BC; ∠ADE=65°, ∠AED=72°: ∠ABC=∠ADE=65° (corresponding angles, DE∥BC); ∠ACB=∠AED=72° (corresponding angles, DE∥BC). So ∠ABC=65° and ∠ACB=72°
   3. Sum of exterior angles of any triangle = 360°: Let interior angles be α, β, γ. Each exterior angle = 180° minus its interior angle. Sum of exteriors = (180−α)+(180−β)+(180−γ) = 540−(α+β+γ) = 540−180 = 360° (angle sum of triangle = 180°)

10. Congruence proofs and applications in real contexts

    1. Plot 1 sides 45, 62, 78; Plot 2 sides 62, 45, 78: The three sides are identical (just listed in different order). By SSS, the triangles are congruent. Yes, their areas are equal (congruent triangles have equal area).
    2. AB=DE, BC=EF, ∠ABC=∠DEF: Two sides and the included angle are equal: ∴ ▵ABC≡▵DEF (SAS). CF = the distance between the non-shared vertices; since the triangles are congruent and share B=E, we have AC=DF (corresponding parts). CF depends on the specific arrangement — state AC=DF by corresponding parts of congruent triangles.
    3. Truss struts 5.2 m, 6.8 m, included angle 74°: SAS guarantees the second truss is congruent to the first (two sides and the included angle equal). The included angle is critical because SSA (two sides and a non-included angle) does not uniquely determine a triangle — two different triangles can be formed with the same two sides and a non-included angle, so the truss dimensions would not be guaranteed to match.