Given a triangle $ABC$ with $AB = AC$, angle $\angle A = 100^o$ and $BD$ bisector of angle $\angle B$. Prove that $$BC = BD + DA.$$
Problem
Source: Indonesia INAMO Shortlist 2008 G10
Tags: geometry, equal segments, isosceles
OlympusHero
26.08.2021 01:07
There's probably a super slick solution but we'll just take the obvious route.
Denote $BD=x$. Then from LoS, we have $AD=\frac{x \sin 20}{\sin 100}$ and $BC=\frac{x \sin 120}{\sin 40}$, so it suffices to show that $\frac{\sin 120}{\sin 40}=1+\frac{\sin 20}{\sin 100}$. This is equivalent to showing $\sin 120 \sin 100 = \sin 40 \sin 100 + \sin 40 \sin 20$. By repeatedly applying Product-To-Sum and $\cos a = -\cos(180-a)$, we get $\cos 20 + \cos 40 = \cos 60 + \cos 40 + \cos 20 - \cos 60$, which is obviously true. All steps are reversible, so we're done.
i3435
26.08.2021 01:24
Let $E$ be on $BC$ so that $BD=BE$. Then by angle chase $ED=EC$, so it suffices to show that $AD=DE$. Let $E'$ be the other point on $BC$ such that $DE=DE'$, then by angle chase $\triangle BAD\cong BE'D$, so $AD=E'D=DE$ as desired.
zerononnatural
26.08.2021 02:19
Let $T$ be on $BC$ such that $BD = BT$. With a simple angle chasing we found that $TD = TC$, so we just need to prove that $TC = TD = DA$, but this is simple since $BADT$ is cyclic, then $\angle TAD = \angle ATD = 20ยบ$. And we are done.