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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 695.
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extend BA to KBC bisectorBD=BKJoin C to KCK=CDsee the KCDA cydric quatrilateralx=25
Problem 695 again:Let the circle through B, C, D cut AC at E.∠EBD = ∠ECD = 27° and so ∠EBC = 90°.Let O be the midpoint of EC.OB = OE = OC and O is the centre of circle BCD∠AOD = ∠EOD = 2∠ECD = 54° = ∠ABDSo A,B,O,D are concyclic.Hencex = ∠DAO = ∠DBO = 63° - ∠OBC = 63° - ∠OCB = 63° - 38° = 25°
Let < ABD be bisected by BEF, E on AC and F on CD. Then BCDE and ABCF are both cyclic so since < EBC = 90, < EDC = 90 = < CAF which shows that AEDF is cyclic. Hence x = < EFD = 25Sumith PeirisMoratuwaSri Lanka
Further we see that E is the incentre of Tr. ABD. The 3 sides of Tr. ECF are the diameters of the 3 cyclic quads in this problem
Problem 695Forms the isosceles triangle AΒΕ with ΒE=ΑΕ and <ABE=<BAE=52.( Point E located under the D ). Then <BEA=76=2.38=2.<BCA. So the point E is circumcenter triangle ABC. But <BAC=180-54-63-38=25, so <ECA=<EAC=52-25=27, (AE=BE=CE) is <DCA=27 then C,D,EIs collinear with <EBD=54-52=2 and <BDC=180-63-38-27=52=<BAE. So the ABDE is Cyclic .Then <EAD=<EBD=2.Therefore x=<DAC=27-2=25.APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE