Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 695.
Wednesday, November 30, 2011
Problem 695: Triangle, Angles, Auxiliary Construction, Congruence, Mind Map, Polya
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extend BA to K
ReplyDeleteBC bisector
BD=BK
Join C to K
CK=CD
see the KCDA cydric quatrilateral
x=25
Problem 695 again:
ReplyDeleteLet 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° = ∠ABD
So A,B,O,D are concyclic.
Hence
x = ∠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.
ReplyDeleteHence x = < EFD = 25
Sumith Peiris
Moratuwa
Sri Lanka
Further we see that E is the incentre of Tr. ABD.
ReplyDeleteThe 3 sides of Tr. ECF are the diameters of the 3 cyclic quads in this problem
Problem 695
ReplyDeleteForms 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,E
Is 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