361. Arhimedes theorem of the broken chord in a circle.
by Virgil Nicula, Oct 23, 2012, 9:07 AM
Theorem of the broken chord (Arhimedes). Let 
, 
on the circle 
, 
the midle of the 
, 
and projection 
of on 
. Prove that 
Proof I (synthetic). Denote the reflection
of the point 
w.r.t. the point . Observe that 
, 
and 
i.e.
. Thus, 
. In conclusion, 
and 
i.e. .
Proof II (synthetic). Denote
. Observe that 
. In conclusion, 
Remarks.
If 
and 
, then the quadrilateral 
is cyclically.
Exists the relation 
(from C.d.p. - Gh. Titeica) 
.
Proof. Denote
for which 
. Observe that 
, 
and 
. Apply the Ptolemy's theorem
in the isosceles trapezoid
: 
.
Proof III (metric).
. From the relation obtain 
.
Proof IV. Let
so thst 
and midpoint 
of 
. It is clear that 
, so 
is cyclic. Now 
, i.e. 
. Hence 
. Since 
, then 
, as desired. We can prove analogously the following extension :
Extension. Let
be a triangle with the circumcircle 
. Denote the point 
for which 
. For a point 
which doesn't contain
the point define the second interserction between the line with the circumcircle of the triangle 
. Prove that 
.
Extension I. Let be the circumcircle of 
. Let 
be a point for which 
. For a point 
(arc) which doesn't contain
the point define the second intersection between the line 
with the circumcircle of 
. Prove that 
.
Extension II. Let be the circumcircle of . Let be the middlepoint of the side ![$ [BC]$](//latex.artofproblemsolving.com/3/5/5/3550468aa97af843ef34b8868728963dec043efe.png)
. For a point (arc) which doesn't contain
the point define the second intersection between the line with the circumcircle of . Prove that 
.
A strong extension . Let be the circumcircle of the triangle 
. Consider a point . For a point (arc) which doesn't contain
the point define the second intersection between the line with the circumcircle of . Prove that 
.
Proof. Let
so that 
. Thus, 
and 
.
Since
obtain that 
. Therefore,
.
Remark.
(the remarkable Archimedes' problem).
, 
on the circle 
, 
the midle of the 
, 
and projection 
of on 
. Prove that 
Proof I (synthetic). Denote the reflection
of the point 
w.r.t. the point . Observe that 
, 
and 
i.e.
. Thus, 
. In conclusion, 
and 
i.e. .Proof II (synthetic). Denote
. Observe that 
. In conclusion, 
Remarks.
If 
and 
, then the quadrilateral 
is cyclically.
Exists the relation 
(from C.d.p. - Gh. Titeica) 
.Proof. Denote
for which 
. Observe that 
, 
and 
. Apply the Ptolemy's theoremin the isosceles trapezoid
: 
.Proof III (metric).
. From the relation obtain 
.Proof IV. Let
so thst 
and midpoint 
of 
. It is clear that 
, so 
is cyclic. Now 
, i.e. 
. Hence 
. Since 
, then 
, as desired. We can prove analogously the following extension :Extension. Let
be a triangle with the circumcircle 
. Denote the point 
for which 
. For a point 
which doesn't containthe point
define the second interserction between the line with the circumcircle of the triangle 
. Prove that 
.Extension I. Let
be the circumcircle of 
. Let 
be a point for which 
. For a point 
(arc) which doesn't containthe point
define the second intersection between the line 
with the circumcircle of 
. Prove that 
.Extension II. Let
be the circumcircle of . Let be the middlepoint of the side ![$ [BC]$](http://latex.artofproblemsolving.com/3/5/5/3550468aa97af843ef34b8868728963dec043efe.png)
. For a point (arc) which doesn't containthe point
define the second intersection between the line with the circumcircle of . Prove that 
.A strong extension . Let
be the circumcircle of the triangle 
. Consider a point . For a point (arc) which doesn't containthe point
define the second intersection between the line with the circumcircle of . Prove that 
.Proof. Let
so that 
. Thus, 
and 
.Since
obtain that 
. Therefore,
.Remark.
(the remarkable Archimedes' problem).This post has been edited 16 times. Last edited by Virgil Nicula, Nov 16, 2015, 1:36 PM
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