(a) Prove that the sum of the angles in a triangle... | WAEC SSCE - General Mathematics - 1990 (Essay)

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(a) Prove that the sum of the angles in a triangle is 2 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.

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(a) The angles of a triangle sum to two right angles.

Let triangle ABC have interior angles at A, B and C. Through C draw a line XY parallel to AB, with X and Y on opposite sides of C.

\(\angle XCA = \angle CAB\) (alternate angles, XY \(\parallel\) AB, transversal CA).
\(\angle YCB = \angle CBA\) (alternate angles, XY \(\parallel\) AB, transversal CB).

The angles on the straight line XY at C add to two right angles:

\[ \angle XCA + \angle ACB + \angle YCB = 180^\circ. \]

Replacing the two alternate angles gives

\[ \angle CAB + \angle ACB + \angle CBA = 180^\circ, \]

so the three interior angles of the triangle sum to two right angles. \(\blacksquare\)

(b) Let \(\angle ACE = \angle ECB = c\) (CE bisects \(\angle ACB\)), and write \(\angle CAB = A\) and \(\angle CBA = B\).

\(\angle CAE = A\). The exterior angle at B, \(\angle CBD\), equals the sum of the two remote interior angles:

\[ \angle CBD = A + \angle ACB = A + 2c. \]

Therefore

\[ \angle CAE + \angle CBD = A + (A + 2c) = 2A + 2c. \tag{1} \]

Now \(\angle CEB\) is the exterior angle of triangle ACE at E, so it equals the sum of the two remote interior angles \(\angle CAE\) and \(\angle ACE\):

\[ \angle CEB = A + c \implies 2\,\angle CEB = 2A + 2c. \tag{2} \]

From (1) and (2),

\[ \angle CAE + \angle CBD = 2\,\angle CEB. \ \blacksquare \]

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Triangles And Polygons

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(a) Prove that the sum of the angles in a triangle is 2 right angles. (b) The side AB of a triangle ABC is produced to a point D. The bisector of ACB cuts A...

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(a) Prove that the sum of the angles in a triangle is 2 right angles. (b) The side AB of a triangle ABC is produced to a point D. The bisector of ACB cuts A...

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