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 2 of 4: Compare the altitudes from E when CF < CG

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In plain words

Two isosceles triangles with equal legs $EF = EC$ have a longer altitude over the shorter base.

CF < CG \implies KF = \frac{CF}{2} < \frac{CG}{2} = LC \implies KE = \sqrt{EF^2 - KF^2} > \sqrt{EC^2 - LC^2} = LE

Detailed analysis

Assume $CF < CG$ . Let $EK \perp DC$ and $EL \perp BC$ be the altitudes from $E$ in the isosceles triangles $ECF$ ( $EF = EC$ ) and $EGC$ ( $EG = EC$ ), so $K$ and $L$ are the midpoints of $CF$ and $CG$ . Then $KF = \dfrac{CF}{2} < \dfrac{CG}{2} = LC$ ; since $EF = EC$ , the Pythagorean theorem in right triangles $EKF$ and $ELC$ gives $KE = \sqrt{EF^2 - KF^2} > \sqrt{EC^2 - LC^2} = LE$ .