Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.1 National olympiad Prove it Romania

Triangle $ABC$ has $\angle A = 90^\circ$ , $\angle B = 30^\circ$ , and $D$ is the foot of the altitude from $A$ . Let $E \in (AD)$ be such that $DE = 3AE$ and let $F$ be the foot of perpendicular from $D$ on $BE$ .

a) Prove that $AF \perp FC$ .

b) Find the measure of the angle $AFB$ .

Solution

a) The hypothesis gives $AC = \frac{1}{2}BC$ and $CD = \frac{1}{2}AC$ , hence $CD = \frac{1}{4}BC$ .
From $\triangle BDE \sim \triangle DFE$ follows $\frac{BD}{DF} = \frac{DE}{FE}$ , that is $\frac{DE}{BD} = \frac{EF}{FD}$ . Now $\frac{DC}{DB} = \frac{1}{3} = \frac{AE}{DE}$ yields $\frac{DE}{BD} = \frac{AE}{DC}$ , so $\frac{AE}{CD} = \frac{EF}{DF}$ .
Now $\widehat{AEF} = 180^\circ - \widehat{DEF} = 180^\circ - (90^\circ - \widehat{ADF}) = 90^\circ + \widehat{ADF} = \widehat{FDC}$ .
Therefore $\triangle AEF \sim \triangle CDF$ (SAS), whence $\widehat{AFE} \equiv \widehat{CFD}$ . This leads to $\widehat{AFC} =$
$\widehat{AFE} + \widehat{EFC} = \widehat{CFD} + \widehat{EFC} = \widehat{EFD} = 90^\circ$ , hence $AF \perp FC$ .

b) Since $\widehat{AFC} = 90^\circ = \widehat{ADC}$ , the quadrilateral $AFDC$ is cyclic, therefore $\widehat{DFC} = \widehat{DAC} = 30^\circ$ . Then $\widehat{AFB} = 360^\circ - \widehat{BFD} - \widehat{DFC} - \widehat{CFA} = 150^\circ$ .

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