Let ABC be a C-right-angled triangle and D a point onto the side AC, K a point onto the segment BD so that <DAK=<AKD=<ABC. Prove that BK=2 CD.
Proof:
Clearly AC>BC and BC is tangent to the circle (ABK), this coming from <AKD=<ABC; let AC intersect this circle second time at E, further to C than A; from <DAK=<ABC it follows that
arc AE=arc KB, i.e. AE=KB and EBKA is an isosceles trapezoid, hence DE=DB. Take O the center of this circle and F midpoint of AE; DEOB is a rhombus ( a kite with OB||DE ), consequently OE||=DB, triangles EOF and DBC are congruent, EF=CD, thus BK=AE=2EF=2CD, done
arc AE=arc KB, i.e. AE=KB and EBKA is an isosceles trapezoid, hence DE=DB. Take O the center of this circle and F midpoint of AE; DEOB is a rhombus ( a kite with OB||DE ), consequently OE||=DB, triangles EOF and DBC are congruent, EF=CD, thus BK=AE=2EF=2CD, done
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