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

Labels:
angle,
congruence,
triangle

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ReplyDeleteextend 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