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 3 of 4: Use the cyclic quadrilateral BCED to compare DK and BL

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In plain words

Inscribed angles in the cyclic quadrilateral $BCED$ make the two right triangles $DEK$ and $BEL$ similar, transferring $KE > LE$ to $DK > BL$ .

\angle EDC = \angle EBC \implies \triangle DEK \sim \triangle BEL \implies (KE > LE \implies DK > BL)

Detailed analysis

Since $BCED$ is cyclic, $\angle EDC = \angle EBC$ , so the right triangles $DEK$ (right-angled at $K$ ) and $BEL$ (right-angled at $L$ ) are similar. Because $KE > LE$ from Step 2, the corresponding legs satisfy $DK > BL$ .