Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.2 National olympiad Prove it

Exercise 2. Let ABC be an isosceles triangle at $A$ , such that $\widehat{B A C}=100^{\circ}$ . Let $\mathrm{D}$ be the point of intersection of (AC) and the bisector of $\widehat{\mathrm{ABC}}$ .

Show that $B C=A D+B D$ .

Solution

Solution to Exercise 2 According to the law of sines, note that $\mathrm{BD} \leqslant \mathrm{BC}$ if and only if $\sin (\widehat{\mathrm{BCD}}) \leqslant$ $\sin (\widehat{\mathrm{BDC}})$ . Since $\widehat{\mathrm{BDC}}=130^{\circ}$ and $\widehat{\mathrm{BCD}}=40^{\circ}$ , we have $\mathrm{BD} \geqslant \mathrm{BC}$ . Let $E$ be the point on $[\mathrm{BC}]$ such that $B D=B E$ , and let $A^{\prime}$ be the symmetric point of $A$ with respect to (AD).

The law of sines then indicates that

$C E=\frac{\sin (\widehat{C D E})}{\sin (\widehat{E C D})} D E=\frac{\sin (\widehat{C D E})}{\sin (\widehat{E C D})} \times \frac{\sin \left(\widehat{E^{\prime} D}\right)}{\sin \left(\widehat{A^{\prime} E D}\right)} A^{\prime} D=\frac{\sin \left(40^{\circ}\right) \sin \left(80^{\circ}\right)}{\sin \left(40^{\circ}\right) \sin \left(80^{\circ}\right)} A D=A D$

We deduce that $B C=C E+B E=A D+B D$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.