Theory question 103
Triangles and Quadrilaterals(a) Prove that the sum of the angles of a triangle is two right angles
(b) The side AB of a triangle ABC is produced to a point D. The bisector of ACB cuts AB at E. Prove that ∠CAE + ∠CBD = 2∠CEB.
Show the model answer
Model answer
(a) Given triangle ABC. Draw a line through A parallel to BC, making angles p (beside B) and q (beside C) at A.
p = ∠B and q = ∠C (alternate angles). p + ∠A + q = 180° (angles on a straight line). So ∠A + ∠B + ∠C = 180° = two right angles.
(b) Let ∠ACE = ∠ECB = c (CE bisects ∠ACB) and ∠CAE = a.
∠CBD is an exterior angle of triangle ABC: ∠CBD = a + 2c.
∠CEB is an exterior angle of triangle AEC: ∠CEB = a + c.
So ∠CAE + ∠CBD = a + a + 2c = 2(a + c) = 2∠CEB.
Explanation
An exterior angle of a triangle equals the sum of the two interior opposite angles.