Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.0 National olympiad Prove it Saudi Arabia

The quadrilateral $ABCD$ has $AD = DC = CB < AB$ and $AB \parallel CD$ . Points $E$ and $F$ lie on the sides $CD$ and $BC$ such that $\widehat{ADE} = \widehat{AEF}$ . Prove that:

a. $4CF \leq CB$ .

b. If $4CF = CB$ , then $AE$ is the angle bisector of $\widehat{DAF}$ .

Solution

a.
$\widehat{FEC} = 180^\circ - \widehat{AEF} - \widehat{DEA}$
$= 180^\circ - \widehat{ADE} - \widehat{DEA} = \widehat{DAE}$
From $AD = DC = CB < AB$ and $AB \parallel CD$ it follows that $\widehat{ADC} = \widehat{DCB}$ , hence triangles $ADE$ and $ECF$ are similar. This yields
$\frac{AD}{EC} = \frac{AE}{EF} = \frac{DE}{CF} . \tag{1}$
This leads to
$AD \cdot CF = EC \cdot DE \leq \frac{1}{4}(EC + DE)^2 = \frac{1}{4} CD^2,$
hence $4CF \leq CB$ , because $AD = DC = CB$ .

b.
If $4CF = CB$ , then the inequality
$EC \cdot DE \leq \frac{1}{4}(EC + DE)^2$
becomes an equality, that is $CE = ED$ .

Relation (1) becomes $\frac{AD}{DE} = \frac{AE}{EF}$ . Since $\widehat{ADE} = \widehat{AEF}$ , triangles $ADE$ and $AEF$ are similar.

Then $\widehat{DAE} = \widehat{EAF}$ , so $AE$ bisects the angle $\widehat{DAF}$ .

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