In a quadrilateral ABCD, angle A =
angle D = 60 degrees, AC = BD, AB = a, AD = 3a, and the length
of CD is greater than the length of AB. AC and BD intersect at
E. Prove that
(2) AE = 3BE; (3) BC is perpendicular to CD; (4)
(Trigonometry is not allowed.)
This entry contributed by Sumith Peiris, Moratuwa, Sri Lanka.