ABC is an isosceles right angled triangle with the right angle at C. Points D and E, equidistant form C, are chosen arbitrarily on AC and BC. Perpendiculars from D and C to AE meet the hypotenuse AB at K an L. Prove that KL = BL.
To see why this is so, extend AC to AF with CF = CD. Triangles ACE and BCF are equal and are obtained from each other with a rotation through 90o. This is then the angle between their corresponding elements. In particular, BF is perpendicular to AE and is therefore parallel DK and CL. This gives two lines AB and AC crossed by three parallel lines, DK, CL, BF. Since CD = CF, so also KL = BL.
