Competition problem — Geometry — Olympiad Maths Prep

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Geometry Difficulty 6.7 National olympiad Prove it Romania

Consider the triangle $ABC$ , with $m(\widehat{A}) > 60^\circ$ and $m(\widehat{C}) > 30^\circ$ . In the half-plane determined by the line $BC$ not containing $A$ , consider points $D$ and $E$ so that $m(\widehat{ABE}) = m(\widehat{CBD}) = 90^\circ$ and $m(\widehat{BAE}) = m(\widehat{BCD}) = 60^\circ$ . Denote $F$ and $H$ the midpoints of the segments $[AE]$ , respectively $[CD]$ , and $G$ the common point of the lines $AC$ and $DE$ . Show that:

a) $\Delta EBD \sim \Delta ABC$ ;

b) $\Delta FGH \equiv \Delta ABC$ .

Solution

a) The hypothesis yields $\Delta ABE \sim \Delta CBD$ , whence $\frac{BE}{BD} = \frac{AB}{CB}$ , that is $\frac{EB}{AB} = \frac{BD}{BC}$ . Since angles $ABC$ and $EBD$ have the same complement, $\widehat{ABC} = \widehat{EBD}$ , hence $\Delta EBD \sim \Delta ABC$ (S.A.S.).

b) From $\Delta EBD \sim \Delta ABC$ follows that $\widehat{ACB} = \widehat{BDE}$ , whence
$m(\widehat{DGC}) = 360^\circ - (m(\widehat{CBD}) + m(\widehat{BDE}) + m(\widehat{BCG})) = 360^\circ - (m(\widehat{CBD}) + m(\widehat{ACB}) + m(\widehat{BCG})) = 90^\circ.$

$[GF]$ is a median in the right triangle $GAE$ , hence $GF = AE/2$ . Since $[AB]$ opposes to an angle of $30^\circ$ in the right triangle $BAE$ , it follows that $AB = \frac{1}{2}AE$ , hence $[AB] = [FG]$ . In the same way, $[BC] = [GH]$ .

From $\triangle FGH \cong \triangle FBH$ (S.S.S.) follows that $m(\widehat{FGH}) = m(\widehat{FBH}) = 90^\circ - m(\widehat{HBD}) - m(\widehat{CBF})$ .

From $m(\widehat{HBD}) = m(\widehat{HDB}) = 30^\circ$ and $m(\widehat{CBF}) = m(\widehat{ABF}) - m(\widehat{ABC}) = 60^\circ - m(\widehat{ABC})$ , we get $m(\widehat{FGH}) = m(\widehat{ABC})$ , so $\triangle FGH \cong \triangle ABC$ (S.A.S.).

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