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<BAC = <BCE, 2|AD| = |ED|

Source: Dutch NMO 2022 p4

November 17, 2022
geometryequal segmentsequal angles

Problem Statement

In triangle ABCABC, the point DD lies on segment ABAB such that CDCD is the angle bisector of angle ∠C\angle C. The perpendicular bisector of segment CDCD intersects the line ABAB in EE. Suppose that ∣BE∣=4|BE| = 4 and ∣AB∣=5|AB| = 5. (a) Prove that ∠BAC=∠BCE\angle BAC = \angle BCE. (b) Prove that 2∣AD∣=∣ED∣2|AD| = |ED|.
[asy] unitsize(1 cm);
pair A, B, C, D, E;
A = (0,0); B = (2,0); C = (1.8,1.8); D = extension(C, incenter(A,B,C), A, B); E = extension((C + D)/2, (C + D)/2 + rotate(90)*(C - D), A, B);
draw((E + (0.5,0))--A--C--B); draw(C--D); draw(interp((C + D)/2,E,-0.3)--interp((C + D)/2,E,1.2));
dot("AA", A, SW); dot("BB", B, S); dot("CC", C, N); dot("DD", D, S); dot("EE", E, S); [/asy]
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