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Vertically Opposite Angles

The side BC...

Question

The side $B C$ of a $△ A B C$ is produced, such that $D$ is on ray $B C$ . The bisector of $∠ A$ meets $B C$ in $∠$ as. Shown in the figure. Prove that $∠ A B C + ∠ A C D = 2 ∠ A L C$ .

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Solution

$\begin{matrix} I n Δ A B L l e t ∠ B A L = a & ∠ B A = b e x t e r i o r ∠ A L C = ∠ B A L + ∠ L B A = a + b I n Δ A B C A L b i sec t ∠ B A C & ∠ A C = a e x t e r i o r ∠ A C D = ∠ B A C + ∠ A B C = 2 a + b = ∠ A B C + ∠ A C o = b + (2 a + b) = 2 a + 2 b = 2 (a + b) = 2 ∠ A L C \end{matrix}$

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The side BC of a $△$ ABC produced, such that D is on ray BC. The bisector of $∠$ A meets BC in L. Prove that $∠$ ABC + $∠$ ACD = 2 $∠$ ALC. [3 MARKS]

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∠ A B C + ∠ A C D = k ∠ A L C

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