## Tuesday, January 25, 2011

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

Geometry Problem
Click the figure below to see the complete problem 574.

Zoom at: Geometry Problem 574

1. Let FM intersection CD be G
Denote 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

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

EH 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

3. name P intersection of AC, DB extended
▲AEP ~ ▲DEP ( CAB = CDB, AEP = DEP )
=> APH = DPH => PH bisector from 572
PH perpendicular to FH

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

5. To Antonio

HEP bisector of CEB, AEC = DEB
AEC + 1/2 CEB = DEB + 1/2 CEB

6. 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.

7. To Antonio,Thanks. H, E, P are never collinear
HEP 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

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

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

Sumith Peiris
Moratuwa
Sri Lanka