Problem 1

Let

\Gamma

be the circumcircle of acute triangle

ABC

. Points

D

and

E

are on segments

AB

and

AC

, respectively, such that

AD=AE

. The perpendicular bisectors of

BD

and

CE

intersect the minor arcs

AB

and

AC

\Gamma

at points

F

and

G

, respectively. Prove that lines

DE

and

FG

are either parallel or they are the same line.

Step 3 of 7: Obtain the second equal distance

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In plain words

The same argument on the other side of the triangle gives the symmetric relation for K.

GC=GE,\qquad \triangle CGE\sim\triangle KAE,\qquad KA=AE.

Detailed analysis

The perpendicular bisector of CE gives GC equal to GE. Using the cyclic points A, C, G, K and the collinearity of A, E, C and G, E, K, the corresponding angle pairs show that triangle CGE is similar to triangle KAE. Thus GC corresponds to GE and CE corresponds to AE, so the equality of the first two sides gives KA equal to AE.