Problem 2

Consider five points

A, B, C, D, E

such that

ABCD

is a parallelogram and

BCED

is a cyclic quadrilateral. Let

\ell

be a line passing through

A

. Suppose that

\ell

intersects the interior of the segment

DC

F

and intersects line

BC

G

. Suppose also that

EF = EG = EC

. Prove that

\ell

is the bisector of angle

DAB

Step 4 of 4: Derive DF > AD and contradict similarity of ADF and GCF

Previous step Next step

In plain words

Subtracting the smaller half-base $KF < LC$ from the larger leg $DK > BL$ forces $DF > BC = AD$ , which directly contradicts $\dfrac{AD}{DF} = \dfrac{CG}{CF} > 1$ .

DF = DK - KF > BL - LC = BC = AD, \qquad \triangle ADF \sim \triangle GCF \implies \frac{AD}{DF} = \frac{CG}{CF} > 1

Detailed analysis

Using $DK > BL$ (Step 3) and $KF < LC$ (Step 2), we get $DF = DK - KF > BL - LC = BC = AD$ , so $\dfrac{AD}{DF} < 1$ . On the other hand, since $AD \parallel GC$ , $\triangle ADF \sim \triangle GCF$ , giving $\dfrac{AD}{DF} = \dfrac{CG}{CF} > 1$ (by the assumption $CF < CG$ ), a contradiction. The case $CF > CG$ gives the exact reverse inequalities $KF > LC$ , $KE < LE$ , $DK < BL$ , $DF < AD$ and $1 < \dfrac{AD}{DF} = \dfrac{CG}{CF} < 1$ , the same contradiction. Therefore $CF = CG$ , and by Step 1 $\ell$ bisects $\angle DAB$ .