BC = AC + AD if <A=2<B, CD angle bisector

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BC = AC + AD if <A=2<B, CD angle bisector

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Source: 2011 Ukraine MO grade VIII P2

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#1 h Jul 14, 2020, 6:45 AM

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In triangle $ABC$ , angle $A$ is twice as large as angle $B, CD$ is the bisector of angle $C$ . Prove that $BC = AC + AD$ .

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#2 h Feb 18, 2021, 2:37 AM

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Construct point $E$ on $BC$ such that $AC = CE$ . We see $\triangle ADC \cong \triangle EDC$ , so $AD = DE$
Let m $\angle DBC = \theta$ . Since m $\angle DEC = 2\theta$ , we have by exterior angle theorem
m $\angle EDB =$ m $\angle EBD = \theta$ , so $DE = EB$ .
$BC = CE + EB = AC + DE = AC + AD$ $\blacksquare$

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