Problem 1

Let

ABCD

be a quadrilateral with all sides equal and

\angle ABC=60^\circ

. Let

\ell

be a line through

D

that does not meet the quadrilateral except at

D

. Let

E,F

be the intersections of

\ell

with

AB,BC

, respectively, and let

M=CE\cap AF

. Prove that

CA^2=CM\cdot CE

Step 3 of 5: Establish the second similarity

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\angle EAC=180^\circ-\angle BAC=120^\circ=\angle ACF\Longrightarrow \triangle EAC\sim\triangle ACF

Detailed analysis

Because $E$ lies on the line $AB$ on the opposite side from $B$ , $\angle EAC=180^\circ-\angle BAC=120^\circ$ . Also triangle $ACD$ is equilateral and $CF$ is parallel to $AD$ , so $\angle ACF=120^\circ$ . Together with $AE/AC=AC/CF$ , the included-angle similarity criterion gives $\triangle EAC\sim\triangle ACF$ .