Geometry Problem

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## Tuesday, January 25, 2011

### Problem 574: Circle, Chord, Secant, Angle Bisector, Perpendicular

Labels:
angle,
angle bisector,
chord,
circle,
perpendicular

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Let FM intersection CD be G

ReplyDeleteDenote the "acute" angles at

A, B, C, D, E, F, G

by the corresponding small letters

a, b, c, d, e, f, g.

b = d = f + c

f = b – c

d = g + (f/2) ,

g = d – (f/2) = b – [(b-c)/2] = (b + c)/2

e + b + c = Pi,

(e/2) = (Pi/2) – [(b+c)/2] = (Pi/2) – g

(e/2) + g = Pi/2 in triangle EHG

So angle at H = 90 degrees

http://img137.imageshack.us/img137/1631/problem574.png

ReplyDeleteEH cut AD at K and BC at L ( see attached picture)

Note that Triangle KED similar to tri. LEB (case AA)

So angle(LKF)=angle(KLF) and triangle KLF is isosceles

And angle bisector FH of isosceles triangle KLF perpendicular to the base KL at H

Peter Tran

name P intersection of AC, DB extended

ReplyDelete▲AEP ~ ▲DEP ( CAB = CDB, AEP = DEP )

=> APH = DPH => PH bisector from 572

PH perpendicular to FH

Problem #574. To c.t.e.o: What is the reason for your statement angle AEP = angle DEP?

ReplyDeleteTo Antonio

ReplyDeleteHEP bisector of CEB, AEC = DEB

AEC + 1/2 CEB = DEB + 1/2 CEB

Problem #574. To c.t.e.o: Thanks for your comment. What is the reason for your statement HEP is the bisector of CEB (why are points H, E, and P collinear?). Thanks.

ReplyDeleteTo Antonio,Thanks. H, E, P are never collinear

ReplyDeleteHEP is the same bisector for vertcales angles

We have to prove bisector of P // to bisector of CBE

AC, DB meet at P1, bisector of CEB meet DP1 at P

bisector of P1 meet CD at E1

In ▲P1CE1 1/2 P1 + P1CE1 + CE1P1 = 180

In E1EPP1 180- CE1P1 + E1EP + E1EP + PBE + 1/2P1=360

from both => 2CE1P1 = 2E1EP => CE1P1 = E1EP

as corresponding angle => P1E1//PE

Problem #574, To c.t.e.o: Thanks.

ReplyDeleteLet G cut BC at X and AD at Y. Then < A = < C. Hence Tr. FXY is isoceles and the result follows.

ReplyDeleteSumith Peiris

Moratuwa

Sri Lanka