(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.
(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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Correct Answer: Option n
Explanation:
(a) Given any triangle ABC
To prove that < A + < B + < C = 180°
Construction: Produce BC to X and draw CP // BA.
Proof: with the lettering \(a_{1} = a_{2}\) (alternate angles);
\(b_{1} = b_{2}\) (corresponding angles)
\(< BCA + < A + < B = 180° ; < ACB + < A + < B = 180°\)
\(\therefore < A + < B + < C = 180°\)
(b)
From the given triangle and with reference to letterings;
\(\hat{a} + \hat{c} + \hat{c} = \hat{b}\) (exterior opposite angles of a triangle)
\(a + c = e\) (exterior opposite angles of a triangle)
\(a + b = a + a + c + c \implies a + b = 2(a + c)\)
= \(2e \text{where e = a + c}\)
i.e \(< CAE + < CBD = 2 < CED\) (Proved)
(a) Given any triangle ABC
To prove that < A + < B + < C = 180°
Construction: Produce BC to X and draw CP // BA.
Proof: with the lettering \(a_{1} = a_{2}\) (alternate angles);
\(b_{1} = b_{2}\) (corresponding angles)
\(< BCA + < A + < B = 180° ; < ACB + < A + < B = 180°\)
\(\therefore < A + < B + < C = 180°\)
(b)
From the given triangle and with reference to letterings;
\(\hat{a} + \hat{c} + \hat{c} = \hat{b}\) (exterior opposite angles of a triangle)
\(a + c = e\) (exterior opposite angles of a triangle)
\(a + b = a + a + c + c \implies a + b = 2(a + c)\)
= \(2e \text{where e = a + c}\)
i.e \(< CAE + < CBD = 2 < CED\) (Proved)