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Parallel lines and angles in triangles

P6 · Geometry · PSLE Maths

Question

Ali is making a trapezium frame ABCD for the class notice board. AB // DC, and diagonal BD is drawn. Given that ∠ABD = 38° and ∠BCD = 112°, find ∠CDB.

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Hint

First use AB // DC to find the co-interior angle ∠ABC, then see which two angles ∠ABC is split into by BD.

Show the answer and step-by-step explanation

Answer: ∠CDB 38 °

  1. Read the diagram — This is trapezium ABCD, with AB // DC. Diagonal BD splits the angle on the right into smaller angles.
  2. Co-interior angles — Since AB // DC, BC is a transversal. So the co-interior angles ∠ABC + ∠BCD = 180°.
  3. Split the angle — At B, ∠ABC = 68°. It is split by BD into 38° and ∠DBC.
  4. Angle sum of triangle — Now look at triangle BCD. Its three interior angles add up to 180°: ∠BCD = 112° and ∠DBC = 30°, so we can find ∠CDB.
  5. Answer — Therefore, ∠CDB = 38°. 📌 Key idea: use parallel lines for co-interior angles, split the larger angle, then use the angle sum of a triangle.

Method: Co-interior angles in parallel lines add up to 180°. Split the larger angle into smaller angles, then use the angle sum of a triangle, 180°, to find the unknown angle.

Practice: 3 similar questions

  1. In trapezium ABCD, AB // DC, and diagonal BD is drawn. Given that ∠ABD = 45° and ∠BCD = 105°, find ∠CDB.
    Answer

    45 ° — Since AB // DC, ∠ABC + ∠BCD = 180°. So ∠ABC = 180° − 105° = 75°. ∠DBC = ∠ABC − ∠ABD = 75° − 45° = 30°. In triangle BCD, ∠CDB = 180° − 105° − 30° = 45°.

  2. Siti drew a trapezium ABCD with AB // DC and joined BD. Given that ∠ABD = 32° and ∠BCD = 116°, find ∠CDB.
    Answer

    32 ° — ∠ABC = 180° − 116° = 64°. ∠DBC = 64° − 32° = 32°. In triangle BCD, ∠CDB = 180° − 116° − 32° = 32°.

  3. Raju drew diagonal BD in trapezium ABCD, where AB // DC. Given that ∠BCD = 124° and ∠CDB = 41°, find ∠ABD.
    Answer

    41 ° — In triangle BCD, ∠DBC = 180° − 124° − 41° = 15°. Since AB // DC, ∠ABC = 180° − 124° = 56°. ∠ABD = ∠ABC − ∠DBC = 56° − 15° = 41°.

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