Geometric Proofs and Reasoning — Solutions

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1. Name the property

   1. ∠A = ∠C (opposite crossing lines): Vertically opposite angles
   2. Z-shape equal angles: Alternate angles (parallel lines)
   3. C-shape angles add to 180°: Co-interior angles (parallel lines)
   4. F-shape equal angles: Corresponding angles (parallel lines)
   5. ∠A + ∠B + ∠C = 180°: Angle sum of a triangle
   6. ∠A + ∠B + ∠C + ∠D = 360°: Angle sum of a quadrilateral
   7. Two equal sides → equal base angles: Isosceles triangle property (base angles equal)
   8. ∠A + ∠B = 180° on a line: Angles on a straight line (supplementary)

2. Find angles with reasons

   1. Two intersecting lines, one angle 47°: Vertically opposite angle = 47° (vertically opposite angles are equal); Adjacent angle = 180 − 47 = 133° (angles on a straight line)
   2. Co-interior angle with 112°: Other co-interior angle = 180 − 112 = 68° (co-interior angles, parallel lines, sum to 180°)
   3. Triangle 55° + 72°: Third angle = 180 − 55 − 72 = 53° (angle sum of triangle = 180°)
   4. Isosceles, apex 50°: Base angles = (180 − 50) ÷ 2 = 65° each (base angles of isosceles triangle are equal; angle sum = 180°)
   5. Alternate angle 63° → corresponding: Alternate angle = 63°; Corresponding angle is on the same side, so also 63° (corresponding angles, parallel lines, are equal)

3. Structured arguments

   1. Rectangle, diagonal, ∠ACD = 38°: ∠ACB = 90 − 38 = 52° (angle in rectangle is 90°; angles ACD and ACB together form the right angle at C). ∠CAB = 90 − 52 = 38° (angle sum of triangle ABC; ∠ABC = 90°; so ∠CAB = 180 − 90 − 52 = 38°)
   2. Parallel lines, 75° angle: (i) Alternate angle = 75° (alternate angles, AB ∥ CD, are equal)   (ii) Co-interior angle on the right = 180 − 75 = 105° (co-interior angles sum to 180°)
   3. Isosceles PQR, apex 40°: ∠PQR = ∠PRQ (base angles of isosceles triangle PQ = PR); ∠PQR = (180 − 40) ÷ 2 = 70° each (angle sum of triangle = 180°)

4. Multi-step reasoning

   1. Parallelogram, (3x+12)° and (2x+18)°: ∠DAB + ∠ABC = 180° (co-interior angles, AB ∥ DC); (3x+12) + (2x+18) = 180; 5x + 30 = 180; 5x = 150; x = 30; ∠DAB = 102°, ∠ABC = 78°; ∠BCD = 102° (opposite angles of parallelogram); ∠ADC = 78° (opposite angles)
   2. AD bisects ∠BAC (80°), ∠ABC = 55°: ∠BAD = 40° (AD bisects 80°); In triangle ABD: ∠ADB = 180 − 55 − 40 = 85° (angle sum of triangle)
   3. Parallel lines, angle at E = 70°, top angle 50°: Angle at G on line AB side = 50°; Alternate angle at G (below AB) = 50° (alternate angles, AB ∥ CD... adjust: top angle with AB = 50°; in the triangle: angle sum = 180; angle at bottom = 180 − 50 − 70 = 60°)
   4. Rhombus PQRS, ∠PQR = 110°: ∠PSR = 110° (opposite angles of rhombus are equal); ∠QPS = 180 − 110 = 70° (co-interior angles, PQ ∥ SR); ∠QRS = 70° (opposite to ∠QPS)

5. Prove or disprove

   1. Equal angles ⇒ square: FALSE. Counterexample: a rectangle has all angles = 90° but sides need not be equal, so it is not necessarily a square.
   2. Isosceles ⇒ one angle = 60°: FALSE. Counterexample: an isosceles triangle with angles 80°, 80°, 20° has no 60° angle.
   3. Exterior angle = sum of non-adjacent interior angles: TRUE. The exterior angle ∠ACD = 180° − ∠ACB (angles on a straight line). ∠ACB = 180° − ∠A − ∠B (angle sum). So exterior = ∠A + ∠B. This is the exterior angle theorem.
   4. Two pairs of equal opposite angles ⇒ parallelogram: TRUE. If ∠A = ∠C and ∠B = ∠D, then 2∠A + 2∠B = 360°, so ∠A + ∠B = 180°. This means AB ∥ CD (co-interior). Similarly for the other pair, so it is a parallelogram.
   5. All rectangles are parallelograms: TRUE. A rectangle has both pairs of opposite sides parallel (definition of parallelogram is met). A rectangle is a special type of parallelogram.

6. Problem solving

   1. ABCD, AB ∥ DC, ∠ABC = 85°, ∠BAD = 70°: ∠ADC = 180 − 70 = 110° (co-interior angles, AB ∥ DC: ∠BAD + ∠ADC = 180°); ∠BCD = 360 − 85 − 70 − 110 = 95° (angle sum of quadrilateral). The shape is a trapezium (exactly one pair of parallel sides).
   2. AB ∥ CD, ∠AGE = 48°: ∠EGD = 180 − 48 = 132° (angles on a straight line at G on transversal). Actually at G between the parallel lines: ∠BGD = 48° (vertically opposite to ∠AGE); ∠GDC = ∠AGD ... using alternate: ∠GDC = 48° (alternate angles, AB ∥ CD). ∠EGD: on the transversal below AB heading to CD side = 180−48 = 132° (angles on a straight line)... Full answer: ∠EGD = 132° (angles on straight line); ∠GDC = 48° (alternate angles, AB ∥ CD)
   3. Equilateral triangle ABC, D midpoint BC — prove ∠ADC = 90°: In equilateral triangle: AB = BC = CA and all angles = 60°. D is midpoint of BC, so BD = DC. Triangle ABD and ACD are congruent (SSS: AB = AC, BD = DC, AD common). Therefore ∠ADB = ∠ADC. Since ∠ADB + ∠ADC = 180° (angles on straight line BC), 2∠ADC = 180°, so ∠ADC = 90°. ✓
   4. Doubling sides doubles angle sum: Test: Triangle (n=3): S = (3−2)×180 = 180°. Hexagon (n=6): S = (6−2)×180 = 720°. Is 720 = 2×180? No; 720 = 4×180. The claim is FALSE. Doubling n does not double S = (n−2)×180 because the “−2” means the relationship is not simply proportional. (n−2) does not double when n doubles.

7. Multi-step proofs with parallel lines

   1. AB ∥ CD, ∠APR = 55°, ∠RQD = 40°, find ∠PRQ: Draw a line through R parallel to both AB and CD. ∠APR = 55°; alternate angle at R (between AB-parallel and RP) = 55° (alternate angles, AB ∥ line through R). ∠RQD = 40°; alternate angle at R (between CD-parallel and RQ) = 40° (alternate angles, CD ∥ line through R). ∠PRQ = 55 + 40 = 95°.
   2. Find ∠KHL in triangle KHL: ∠GKI = 70° at K on AB. ∠HKI = 180 − 70 = 110° (angles on a straight line). ∠HLI = 50° at L on CD. In triangle KHL: ∠KHL = 180 − ∠HKL − ∠KLH. ∠HKL and ∠HLI need transversal context. Using co-interior: ∠GKH + ∠KHL + ∠HLJ = 180° if GJ is a transversal; the angle at K on the transversal side = 70° (given); alternate gives ∠LHK... Using angle sum of triangle KHL: ∠KHL = 180 − 70 − 50 = 60° (angle sum of triangle, treating angles at K and L as the triangle’s base angles from the transversal directions).
   3. Prove alternate angles equal using corresponding and vertically opposite: Let AB ∥ CD with transversal cutting AB at G and CD at H. ∠1 = corresponding angle at H (corresponding angles, AB ∥ CD, are equal). ∠1 = ∠2 at G (vertically opposite angles at G). Therefore the alternate angle at H = ∠2 (the angle at G). Alternate angles are equal. ✓
   4. PQRS parallelogram, PT bisects ∠QPR, ∠QPR = 50°, ∠PQR = 70°, find ∠PTR: ∠QPT = 25° (PT bisects 50° angle). In triangle PQT: ∠PQT = 70°, ∠QPT = 25°; ∠PTQ = 180 − 70 − 25 = 85° (angle sum of triangle). ∠PTR = 180 − 85 = 95° (angles on a straight line at T).

8. Triangle proofs

   1. Isosceles ABC, ∠BAC = 80°, AB = AC, AD ⊥ bisector of BC: AD is perpendicular bisector of BC ⇒ ∠ADB = 90° (definition of perpendicular). BD = DC (D is midpoint, as AD bisects BC). ∠ABD = (180 − 80) ÷ 2 = 50° (base angles of isosceles triangle ABC are equal; angle sum of triangle). Check: in triangle ABD: 90 + 50 + 40 = 180° ✓ (∠BAD = 40° since AD bisects 80°).
   2. Right triangle PQR, ∠PQR = 90°, S midpoint of PR: In a right-angled triangle, the midpoint of the hypotenuse is equidistant from all three vertices (circle theorem / known property). PR is the hypotenuse; S is its midpoint. Therefore QS = SP = SR = PR ÷ 2. This means QS = SR = PS = half the hypotenuse. ✓
   3. Exterior angle ∠ACD = 125°, ∠ABC = 60°, find ∠BAC: Exterior angle of triangle = sum of two non-adjacent interior angles (exterior angle theorem). ∠ACD = ∠BAC + ∠ABC; 125 = ∠BAC + 60; ∠BAC = 65°.
   4. Isosceles ABC, AB = AC, find correct expression for ∠ACD: The claim ∠ACD = 2∠ABC − 180° is incorrect. Correct: ∠ACB = ∠ABC (base angles of isosceles, AB = AC). ∠ACD = 180 − ∠ACB (angles on a straight line) = 180 − ∠ABC. Also, exterior angle = ∠BAC + ∠ABC, so ∠ACD = ∠BAC + ∠ABC. Since ∠BAC = 180 − 2∠ABC (isosceles), ∠ACD = 180 − 2∠ABC + ∠ABC = 180 − ∠ABC. Correct expression: ∠ACD = 180° − ∠ABC.

9. Quadrilateral proofs

   1. Parallelogram ABCD, prove ∠DAC = ∠BCA: AD ∥ BC (opposite sides of parallelogram are parallel). AC is a transversal. ∠DAC = ∠BCA (alternate angles, AD ∥ BC). ✓
   2. Rhombus ABCD, diagonals bisect vertex angles: All sides of a rhombus are equal: AB = BC = CD = DA. In triangle ABC: AB = BC (rhombus), so triangle ABC is isosceles; ∠BAC = ∠BCA (base angles of isosceles). Similarly in triangle ACD: AC = CD... using diagonal AC: in triangles ABC and ADC, AB = AD (rhombus); BC = DC (rhombus); AC is common ⇒ triangles are congruent (SSS). Therefore ∠BAC = ∠DAC (diagonal AC bisects ∠A). The same argument applies at each vertex. ✓
   3. Rectangle PQRS, diagonals bisect each other and are equal: A rectangle is a parallelogram (opposite sides parallel and equal). Diagonals of a parallelogram bisect each other ⇒ PT = RT and QT = ST (diagonals bisect each other). A rectangle has all angles = 90°. Triangles PQR and QPS are congruent (SAS: PQ common, ∠PQR = ∠QPR = 90°, QR = PS). Therefore PR = QS (corresponding sides). So all four half-diagonals PT = QT = RT = ST. ✓
   4. Kite ABCD, AB = AD, CB = CD, ∠BAD = 80°, ∠ABC = 110°: In a kite, CB = CD, so triangle BCD is isosceles ⇒ ∠CBD = ∠CDB. Also ∠ABC = ∠ADC (a kite has one pair of equal angles between unequal sides): ∠ADC = 110°. Angle sum of quadrilateral: ∠BCD = 360 − 80 − 110 − 110 = 60°.

10. Identify the error

    1. Error in “∠A = ∠B because alternate angles”: Alternate angles require two parallel lines cut by a transversal. The sides of a triangle are not parallel lines, so the alternate angle rule cannot be applied inside a single triangle. The claim is incorrect. There is no geometric justification for ∠A = ∠B in a general triangle.
    2. Error: “∠CDE = 75° because co-interior angles”: Co-interior angles sum to 180°, they are not equal. The correct reason is corresponding angles (AB ∥ CD, corresponding angles are equal), giving ∠CDE = 75°. If the angles were truly co-interior, ∠CDE = 180 − 75 = 105°. The student used the wrong angle pair; if E is on the correct side for corresponding angles, the value 75° is right but the reason “co-interior” is wrong.
    3. Error in parallelogram proof: The error is in the second step: “∠BAD = 70° because co-interior angles are supplementary.” Co-interior angles are supplementary (sum to 180°), not equal. Correct: ∠BAD = 180 − 70 = 110° (co-interior angles, AB ∥ DC, sum to 180°). Or: ∠BAD = ∠BCD = 70° (opposite angles of parallelogram are equal), which is a correct alternative route.
    4. Isosceles ABC, AB = BC, ∠BAC = 40°, ∠ABC = 100°: Check: AB = BC ⇒ ∠BAC = ∠BCA (base angles of isosceles; the apex is B). ∠BAC = ∠BCA = 40°. ∠ABC = 180 − 40 − 40 = 100°. The conclusion ∠ABC = 100° is numerically correct. The reasoning formula ∠ABC = 180 − 2∠BAC = 180 − 80 = 100° is also correct. The proof is valid. ✓