Let acute-angled triangle
, and let
be the circumscribed circle. Points
and
in the parts
and
corresponding order
.
The perpendicular bisector of
and
intersecting the small arches
and
the
in points
and
respectively.
Prove that the lines
and
are either parallel or are identical.
Suppose that
intersects the circumscribed circle of
a point
.
We observe that
.
Thus the triangle
is isosceles with
.
Similarly if
intersects circumcircle of
at
, then
.
We will prove that
, ie
.
We note that
, while
(
is the core of
the circle with center
and radius
.
So it suffices to prove that
, that
.
But we know that
, by definition
.
So long
.
We observe that the quadrilateral
is recordable and
therefore the
is the bisector
.
If it
was not equilateral, then since
it belongs to the bisector
and
, should it
be writable (south pole theorem) inappropriate, since the circumscribed circle
is the circumscribed circle
and
does not belong to this circle.
So it
is isosceles with
and the requested follow.
![ABC ABC](https://mathematica.gr/forum/ext/geomar/texintegr/latexrender/pictures/902fbdd2b1df0c4f70b4a5d23525e932.png)
![\Gamma \Gamma](https://mathematica.gr/forum/ext/geomar/texintegr/latexrender/pictures/07710b5c43702a8bb7b9104eacc6ba71.png)
![D D](https://mathematica.gr/forum/ext/geomar/texintegr/latexrender/pictures/f623e75af30e62bbd73d6df5b50bb7b5.png)
![IS IS](https://mathematica.gr/forum/ext/geomar/texintegr/latexrender/pictures/3a3ea00cfc35332cedf6e5e9a32e94da.png)
![FROM FROM](https://mathematica.gr/forum/ext/geomar/texintegr/latexrender/pictures/b86fc6b051f63d73de262d4c34e3a0a9.png)
![AM AM](https://mathematica.gr/forum/ext/geomar/texintegr/latexrender/pictures/4144e097d2fa7a491cec2a7a4322f2bc.png)
![AD AE AD AE](https://mathematica.gr/forum/ext/geomar/texintegr/latexrender/pictures/b9d4cb850d8bfc6f9fa1b3aa94c2f42b.png)
The perpendicular bisector of
![BD BD](https://mathematica.gr/forum/ext/geomar/texintegr/latexrender/pictures/87a47565be4714701a8bc2354cbaea36.png)
![THIS THIS](https://mathematica.gr/forum/ext/geomar/texintegr/latexrender/pictures/7a86131338bf955e0a56311f264aa6aa.png)
![FROM FROM](https://mathematica.gr/forum/ext/geomar/texintegr/latexrender/pictures/b86fc6b051f63d73de262d4c34e3a0a9.png)
![AM AM](https://mathematica.gr/forum/ext/geomar/texintegr/latexrender/pictures/4144e097d2fa7a491cec2a7a4322f2bc.png)
![\Gamma \Gamma](https://mathematica.gr/forum/ext/geomar/texintegr/latexrender/pictures/07710b5c43702a8bb7b9104eacc6ba71.png)
![F F](https://mathematica.gr/forum/ext/geomar/texintegr/latexrender/pictures/800618943025315f869e4e1f09471012.png)
![G G](https://mathematica.gr/forum/ext/geomar/texintegr/latexrender/pictures/dfcf28d0734569a6a693bc8194de62bf.png)
Prove that the lines
![FROM FROM](https://mathematica.gr/forum/ext/geomar/texintegr/latexrender/pictures/3a52f3c22ed6fcde5bf696a6c02c9e73.png)
![FG FG](https://mathematica.gr/forum/ext/geomar/texintegr/latexrender/pictures/2a5271c118492b7bb2274dd278a033ba.png)
Solution
Suppose that
![FD FD](https://mathematica.gr/forum/ext/geomar/texintegr/latexrender/pictures/a8fa6b553b655657f943cb8fd85859d1.png)
![ABC ABC](https://mathematica.gr/forum/ext/geomar/texintegr/latexrender/pictures/902fbdd2b1df0c4f70b4a5d23525e932.png)
![The The](https://mathematica.gr/forum/ext/geomar/texintegr/latexrender/pictures/d20caec3b48a1eef164cb4ca81ba2587.png)
We observe that
![\widehat{ADL}=\widehat{FDB}=\widehat{FBD}=\widehat{FBA}=\widehat{FLA}=\widehat{ALD} \widehat{ADL}=\widehat{FDB}=\widehat{FBD}=\widehat{FBA}=\widehat{FLA}=\widehat{ALD}](https://mathematica.gr/forum/ext/geomar/texintegr/latexrender/pictures/d48671406cc130aaa69f130c7f5c1abf.png)
Thus the triangle
![LAD LAD](https://mathematica.gr/forum/ext/geomar/texintegr/latexrender/pictures/902341c595d016d1b2ebb0feb5c624b1.png)
![AL = AD = AE AL = AD = AE](https://mathematica.gr/forum/ext/geomar/texintegr/latexrender/pictures/8bc5c3a8f3423e555e4f3148e713e647.png)
Similarly if
![GIVE GIVE](https://mathematica.gr/forum/ext/geomar/texintegr/latexrender/pictures/f803729628adf4199f224c2a225038e9.png)
![ABC ABC](https://mathematica.gr/forum/ext/geomar/texintegr/latexrender/pictures/902fbdd2b1df0c4f70b4a5d23525e932.png)
![K K](https://mathematica.gr/forum/ext/geomar/texintegr/latexrender/pictures/a5f3c6a11b03839d46af9fb43c97c188.png)
![AK AL = AE = AD = AK AL = AE = AD =](https://mathematica.gr/forum/ext/geomar/texintegr/latexrender/pictures/41b96aa002c0fa1c4e342716a9b4ebc8.png)
We will prove that
![FG // THE FG // THE](https://mathematica.gr/forum/ext/geomar/texintegr/latexrender/pictures/1d2bd2164b3c23a2b14bc2bc465a14b0.png)
![\widehat{LFG}=\widehat{LDE} \widehat{LFG}=\widehat{LDE}](https://mathematica.gr/forum/ext/geomar/texintegr/latexrender/pictures/d0e4c8d3317c79b3dc7b607859602236.png)
We note that
![\widehat{LFG}=\widehat{LAG} \widehat{LFG}=\widehat{LAG}](https://mathematica.gr/forum/ext/geomar/texintegr/latexrender/pictures/ae0865f30b1bab7f955ea18245d631e4.png)
![\widehat{LDE}=\dfrac{\widehat{LAE}}{2} \widehat{LDE}=\dfrac{\widehat{LAE}}{2}](https://mathematica.gr/forum/ext/geomar/texintegr/latexrender/pictures/6719030095e47222e3383064da1e8365.png)
![\widehat{LAE} \widehat{LAE}](https://mathematica.gr/forum/ext/geomar/texintegr/latexrender/pictures/680c2f5a1205a3b4c787efea9a1f72ee.png)
![\widehat{LDE} \widehat{LDE}](https://mathematica.gr/forum/ext/geomar/texintegr/latexrender/pictures/f8b0a882e856b1c508ea82d3061e342d.png)
![A A](https://mathematica.gr/forum/ext/geomar/texintegr/latexrender/pictures/7fc56270e7a70fa81a5935b72eacbe29.png)
![AK AL = AE = AD = AK AL = AE = AD =](https://mathematica.gr/forum/ext/geomar/texintegr/latexrender/pictures/41b96aa002c0fa1c4e342716a9b4ebc8.png)
So it suffices to prove that
![\widehat{LAG}=\widehat{CAG} \widehat{LAG}=\widehat{CAG}](https://mathematica.gr/forum/ext/geomar/texintegr/latexrender/pictures/9b72c08cee3a898c6f6b35191d38a18c.png)
![GL=GC GL=GC](https://mathematica.gr/forum/ext/geomar/texintegr/latexrender/pictures/c304860bde2e8331c35dd7e0ce5809d6.png)
But we know that
![GC = GE GC = GE](https://mathematica.gr/forum/ext/geomar/texintegr/latexrender/pictures/ccb73ae157263535a9febf061ceccd27.png)
![G G](https://mathematica.gr/forum/ext/geomar/texintegr/latexrender/pictures/dfcf28d0734569a6a693bc8194de62bf.png)
So long
![GL=GE GL=GE](https://mathematica.gr/forum/ext/geomar/texintegr/latexrender/pictures/7998779965ea967eb8347d3a89412cae.png)
We observe that the quadrilateral
![KALG KALG](https://mathematica.gr/forum/ext/geomar/texintegr/latexrender/pictures/595369695548dbc8b9b02b44aaecfca6.png)
![AND AL = AND AL =](https://mathematica.gr/forum/ext/geomar/texintegr/latexrender/pictures/6de92c349cd94df436ca7178b10ea8ea.png)
![AG AG](https://mathematica.gr/forum/ext/geomar/texintegr/latexrender/pictures/6b712da7b7ccf80851beb06de6c32e6c.png)
![\widehat{LGE} \widehat{LGE}](https://mathematica.gr/forum/ext/geomar/texintegr/latexrender/pictures/016d88f7d66cd4efc182749557d358ac.png)
If it
![LGE LGE](https://mathematica.gr/forum/ext/geomar/texintegr/latexrender/pictures/b930e96679130b90a8131af333e67723.png)
![A A](https://mathematica.gr/forum/ext/geomar/texintegr/latexrender/pictures/7fc56270e7a70fa81a5935b72eacbe29.png)
![\widehat{LGE} \widehat{LGE}](https://mathematica.gr/forum/ext/geomar/texintegr/latexrender/pictures/016d88f7d66cd4efc182749557d358ac.png)
![AE = AL AE = AL](https://mathematica.gr/forum/ext/geomar/texintegr/latexrender/pictures/381a2c1e9dfb18cf112aa0d0efceba78.png)
![glaer glaer](https://mathematica.gr/forum/ext/geomar/texintegr/latexrender/pictures/90710ec6c985959363189165f386b0ef.png)
![Alarg Alarg](https://mathematica.gr/forum/ext/geomar/texintegr/latexrender/pictures/1204a5c2ac4e8891367b2b2c03f72bb8.png)
![ABC ABC](https://mathematica.gr/forum/ext/geomar/texintegr/latexrender/pictures/902fbdd2b1df0c4f70b4a5d23525e932.png)
![IS IS](https://mathematica.gr/forum/ext/geomar/texintegr/latexrender/pictures/3a3ea00cfc35332cedf6e5e9a32e94da.png)
So it
![LGE LGE](https://mathematica.gr/forum/ext/geomar/texintegr/latexrender/pictures/b930e96679130b90a8131af333e67723.png)
![GL=GE GL=GE](https://mathematica.gr/forum/ext/geomar/texintegr/latexrender/pictures/7998779965ea967eb8347d3a89412cae.png)
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