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Difference between revisions of "Dao Thanh Oai geometric results"

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(→‎Van Aubel's theorem)
 
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2. Prove that <math>\triangle BSC \sim \triangle ASD.</math>
 
2. Prove that <math>\triangle BSC \sim \triangle ASD.</math>
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3. Let points <math>S_0</math> and <math>S_1</math> be the points symmetric <math>S</math> with respect <math>BC</math> and <math>AD, P_0</math> and <math>P_1</math> be the points symmetric <math>P</math> with respect <math>AB</math> and <math>CD.</math>
 +
 +
Prove that <math>S_0S_1 \perp P_0P_1</math> and <math>\frac {P_0P_1}{S_0S_1} = \frac{|\cot \alpha + \cot \beta|}{2}.</math>
  
 
<i><b>Proof</b></i>
 
<i><b>Proof</b></i>
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It is trivial that <math>t(T(P)) = P.</math>
 
It is trivial that <math>t(T(P)) = P.</math>
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[[File:Dao ginma3.png|300px|right]]
 
It is known ([[ Spiral similarity | Superposition of two spiral similarities]]) that <math>B' = t(T(A))</math> is the point with properties  
 
It is known ([[ Spiral similarity | Superposition of two spiral similarities]]) that <math>B' = t(T(A))</math> is the point with properties  
 
<cmath>AP = BP, \angle APB = 2 \alpha \implies</cmath>
 
<cmath>AP = BP, \angle APB = 2 \alpha \implies</cmath>
 
<cmath>B' = B, \triangle BSC \sim \triangle ASD.</cmath>
 
<cmath>B' = B, \triangle BSC \sim \triangle ASD.</cmath>
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 +
3. <math>PM = MP_1, PM' = M'P_0 \implies</math>
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<cmath>P_0P_1 = 2 M'M, P_0P_1 || M'M.</cmath>
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<cmath>E= SS_1 \cap AB, F = SS_0 \cap CD \implies</cmath>
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<cmath>S_0S_1 = 2 FE, S_0S_1 || FE.</cmath>
 +
<cmath>2 \vec {MM'} = \vec {DA} + \vec {CB}, \vec {EF} = \vec {ES} + \vec {SF},</cmath>
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<cmath>|\vec {ES}| \cdot (\cot \alpha + \cot \beta) = |\vec {DA}|,|\vec {SF}| \cdot (\cot \alpha + \cot \beta) = |\vec {CB}|,</cmath>
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<cmath>ES \perp AD, SF \perp CB \implies EF \perp MM', S_0S_1 \perp P_0P_1,</cmath>
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<cmath>\frac {P_0P_1}{S_0S_1} = \frac{|\cot \alpha + \cot \beta|}{2}.</cmath>
 +
Note: If superposition of two spiral similarities is possible, the result is valid even for positions of point <math>P</math> outside the quadrilateral and for a non-convex quadrilateral.
  
 
==Bottema's theorem==
 
==Bottema's theorem==
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We use [[Dao Thanh Oai geometric results | Dao bisectors theorem]] for quadrilateral <math>ABCD</math> with <math>\alpha = 90^\circ, \beta = 45^\circ</math> and get existence given triangle <math>\triangle SCD</math> with need properties.
 
We use [[Dao Thanh Oai geometric results | Dao bisectors theorem]] for quadrilateral <math>ABCD</math> with <math>\alpha = 90^\circ, \beta = 45^\circ</math> and get existence given triangle <math>\triangle SCD</math> with need properties.
 +
==Napoleon's theorem==
 +
[[File:Napoeon Dao.png|300px|right]]
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[[File:Napoleon Dao Inn.png|300px|right]]
 +
Let isosceles triangles with an angle of 120 degrees at the apex be constructed on the sides of an arbitrary triangle in the outer direction. The triangle with vertices at the apex those triangles names outer Napoleon triangle. Napoleon's theorem states that it is the equilateral triangle.
 +
 +
Let triangle <math>\triangle SCD</math> be given. Let triangles <math>\triangle SDA, \triangle SCB, \triangle CDP</math> be the isosceles triangles with angles <math>120^\circ</math> (see diagram).
 +
 +
Prove that <math>\triangle ABP</math> is the equilateral triangle.
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 +
<i><b>Proof</b></i>
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 +
For given point <math>S</math> one can find points <math>A(B)</math> using rotation point <math>S</math> around <math>D(C)</math> at the <math>30^\circ</math> in counterclockwise (clockwise) direction and homothety with coefficient <math>\frac{1}{\sqrt{3}}.</math>
 +
 +
We use [[Dao Thanh Oai geometric results | Dao bisectors theorem]] for quadrilateral <math>ABCD</math> with <math>\alpha = 30^\circ, \beta = 120^\circ</math> and get <math>AP = BP, \angle APB = 60^\circ.</math>
 +
 +
Note: Napoleon's theorem can also be proved for the inner triangle using the method of a pair of spiral symmetries (see diagram).
 +
 +
==Finsler - Hadwiger theorem==
 +
[[File:Finsler theorem.png|300px|right]]
 +
Let two squares with common vertex be given.  The theorem states that the quadrilateral of midpoints is (the Finsler–Hadwiger) square.
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 +
Let <math>AD</math> and <math>BC</math> be diagonals of given squares with common vertex <math>S.</math> Let <math>M'</math> and <math>M</math> be the midpoints of <math>AB</math> and <math>CD</math> respectively, and let <math>O</math> and <math>O'</math> be the centers of these squares.
 +
 +
Prove that <math>OMO'M'</math> is square.
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<i><b>Proof</b></i>
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Quadrilateral of midpoints is the parallelogram <math>OMO'M'.</math>
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 +
We use [[Dao Thanh Oai geometric results | Dao bisectors theorem]] for quadrilateral <math>ABCD</math> with <math>\alpha = \beta = 45^\circ</math> and get
 +
<cmath>\frac {M'M}{O'O} = \frac{\cot \alpha + \cot \beta}{2} = 1, M'M \perp OO' \implies</cmath>
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<math>OMO'M'</math> is the square.
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==Brahmagupta's theorem==
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[[File:Brachmagupta Dao.png|300px|right]]
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Let a cyclic quadrilateral <math>ABCD</math> with <math>AC \perp BD</math> be given, <math>S = AC \cap BD, M</math> be the midpoint <math>CD.</math>
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Prove that <math>MS \perp AB</math> (Brahmagupta's theorem).
 +
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<i><b>Proof</b></i>
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<math>\angle ADB = \angle ACB = \alpha, \angle DAC = \angle DBC = \beta</math>
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<math>\beta = 90^\circ - \alpha, P</math> is circumcenter <math>\odot ABCD,</math>
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<cmath>\angle PDM = 90^\circ - \beta = \alpha \implies Q = M.</cmath>
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 +
We use [[Dao Thanh Oai geometric results | Dao bisectors theorem]] for quadrilateral <math>ABCD</math> with <math>\alpha + \beta = 90^\circ</math> and get
 +
<cmath>QS \perp AB \implies MS \perp AB.</cmath>
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==Van Aubel's theorem==
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[[File:Van Aubel vectors.png|300px|right]]
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[[File:Van Aubel circle.png|300px|right]]
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Let squares <math>AFDF', AHBH',  BECE', CGDG'</math> constructed on the sides of a quadrilateral <math>ABCD</math> be given.
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Let <math>M, M'</math> be the midpoints <math>AC</math> and <math>BD,</math>
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<cmath>K = FE \cap H'G',K' = HG \cap E'F',</cmath>
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<cmath>L = EF \cap HG, L' = F'E' \cap H'G'.</cmath> Notation is shown in the diagram.
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1. Prove that  <math>FE = GH, FE \perp GH,</math>
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<cmath>F'E' = G'H', F'E' \perp G'H'.</cmath>
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2. Prove that <math>MM'</math> and <math>KK'</math> are the perpendicular diameters of the circle <math>\omega</math> centered at centroid <math>O</math> of <math>ABCD</math> and points <math>L</math> and <math>L'</math> lies on this circle.
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<i><b>Proof</b></i>
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Denote the centers of given squares <math>I, J, I', J',</math>
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<cmath>a =\vec {DI}, a' = \vec IF, b = \vec AI, b' = \vec JH,</cmath>
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<cmath>c = \vec BI', c' = \vec I'E, d = \vec CJ', d' = \vec J'G,</cmath>
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<math>O -</math> the centroid of <math>ABCD.</math>
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It is evident that <math>\vec a + \vec b + \vec c + \vec d = 0.</math>
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Vectors with sign ' are the rotations of the vectors without this sign in counterclockwise direction, so <math>|\vec a| = |\vec {a'}|,...</math> and
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<cmath>\vec {a'} + \vec b' + \vec c' + \vec d' = 0.</cmath>
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<cmath>\vec {OA} + \vec {OB} + \vec {OC} + \vec {OD} = 0 \implies \vec {OA} = \vec a + \frac{\vec d - \vec b}{2},</cmath>
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<cmath>\vec {OB} = \vec b + \frac{\vec a - \vec c}{2},\vec {OC} = \vec c + \frac{\vec b - \vec d}{2}, \vec {OD} = \vec d + \frac{\vec c - \vec a}{2}.</cmath>
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1. <math>\vec {OF} = \vec {OD} + \vec a +\vec {a'} = \frac{\vec d - \vec b}{2} +\vec {a'}, \vec {OF'} = \vec {OF} - 2 \vec {a'} = \frac{\vec d - \vec b}{2} - \vec {a'}.</math>
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<cmath>\vec {OE} = \frac{\vec b - \vec d}{2} +\vec {c'}, \vec {OE'} = \frac{\vec b - \vec d}{2} - \vec {c'}.</cmath>
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<math>\vec {EF} = \vec {OF} - \vec {OE} = \vec d - \vec b +\vec {a'} - \vec {c'}.</math>
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Similarly, <math>\vec {E'F'} = \vec d - \vec b +\vec {c'} - \vec {a'}, \vec {HG} = \vec c - \vec a +\vec {d'} - \vec {b'}, \vec {H'G'} = \vec a - \vec c +\vec {d'} - \vec {b'}.</math>
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 +
<math>\vec {a'}</math> and <math>-\vec a</math> are the results of rotation <math>\vec {a}</math> and <math>\vec {a'}</math> in counterclockwise direction at <math>90^\circ</math> therefore <math>\vec {HG}</math> is the result of rotation <math>\vec {EF}</math> in counterclockwise direction at <math>90^\circ \implies</math>
 +
<cmath>EF = HG, EF \perp HG.</cmath>
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Similarly about <math>\vec {H'G'}</math> and <math>\vec {E'F'}.</math>
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2. <math>\vec OM = \frac{\vec {OA} + \vec {OC}}{2} = \frac{\vec a + \vec c}{2}, </math>
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<cmath>\vec OM' = \frac{\vec {OB} + \vec {OD}}{2} = \frac{\vec b + \vec d}{2} = - \vec {OM},</cmath>
 +
<cmath>\vec {OK} = \vec {OF} + \mu \vec {EF} = \vec {OH' } + \nu \vec {G'H'},
 +
\vec {OF} = \frac{\vec d - \vec b}{2} + \vec {a'}, \vec {OH' }= \frac{\vec a - \vec c}{2} - \vec {b'},</cmath>
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<cmath>\vec {OK} = \frac{\vec a' + \vec c'}{2}.</cmath>
 +
Similarly,  <math>\vec {OK'} = \frac{\vec b' + \vec d'}{2} = -\vec{OK} \implies OM = OK = OM' = OK', OM \perp OK.</math>
 +
<cmath>\angle KLK' = \angle ELH = 90^\circ \implies L \in \omega.</cmath>
 +
<cmath>\angle KL'K' = \angle E'L'H' = 90^\circ \implies L' \in \omega.</cmath>

Latest revision as of 02:41, 2 September 2025

Dao Thanh Oai was born in Vietnam in 1986. He is an engineer with many innovative solutions for Vietnam Electricity and mathematician with a large number of remarkable discoveries in classical geometry. Some of his results are shown and proven below.

Page made by vladimir.shelomovskii@gmail.com, vvsss

Dao bisectors theorem

Dao ginma1.png
Dao ginma2.png

Let a convex quadrilateral $ABCD$ be given. Let $\ell', M'$ and $\ell, M$ be the bisector and the midpoint of $AB$ and $CD,$ respectively. Let $\ell'$ intersect $\ell$ at the point $P$ inside $ABCD.$ Denote $\angle APM' = \alpha, \angle MPD = \beta.$

Let point $S$ be the point inside $ABCD$ such that $\angle SDA = \alpha, \angle SAD = \beta.$

Let $Q$ be the point at ray $PM$ such that $\angle PDQ = \alpha.$ Define $Q'$ similarly.

1. Prove that $SQ \perp AB, SQ' \perp CD, SQ = PQ'.$

2. Prove that $\triangle BSC \sim \triangle ASD.$

3. Let points $S_0$ and $S_1$ be the points symmetric $S$ with respect $BC$ and $AD, P_0$ and $P_1$ be the points symmetric $P$ with respect $AB$ and $CD.$

Prove that $S_0S_1 \perp P_0P_1$ and $\frac {P_0P_1}{S_0S_1} = \frac{|\cot \alpha + \cot \beta|}{2}.$

Proof

1. $\angle SAD = \angle QPD, \angle SDA = \angle QDP \implies$ \[\triangle DAS \sim \triangle DPQ.\]

The spiral similarity taking $A$ to $S$ and $P$ to $Q$ has center $D$ and angle $\alpha.$ Therefore spiral similarity taking $AP$ to $SQ$ and $P$ to $Q$ has the same center $D$ and angle $\alpha.$

$\angle APM' = \alpha,$ so $AP$ maps into segment parallel $PM' \implies SQ \perp AB.$

2. Let $T$ be the spiral similarity centered at $D$ with angle $\alpha$ and coefficient $\frac {1}{|k|}, S = T(A), \frac {AD}{SD} = k.$

Let $t$ be spiral similarity centered at $C$ with angle $\alpha$ and coefficient $k, B = t(S), \frac {BC}{SC} = k.$

It is trivial that $t(T(P)) = P.$

Dao ginma3.png

It is known ( Superposition of two spiral similarities) that $B' = t(T(A))$ is the point with properties \[AP = BP, \angle APB = 2 \alpha \implies\] \[B' = B, \triangle BSC \sim \triangle ASD.\]

3. $PM = MP_1, PM' = M'P_0 \implies$ \[P_0P_1 = 2 M'M, P_0P_1 || M'M.\] \[E= SS_1 \cap AB, F = SS_0 \cap CD \implies\] \[S_0S_1 = 2 FE, S_0S_1 || FE.\] \[2 \vec {MM'} = \vec {DA} + \vec {CB}, \vec {EF} = \vec {ES} + \vec {SF},\] \[|\vec {ES}| \cdot (\cot \alpha + \cot \beta) = |\vec {DA}|,|\vec {SF}| \cdot (\cot \alpha + \cot \beta) = |\vec {CB}|,\] \[ES \perp AD, SF \perp CB \implies EF \perp MM', S_0S_1 \perp P_0P_1,\] \[\frac {P_0P_1}{S_0S_1} = \frac{|\cot \alpha + \cot \beta|}{2}.\] Note: If superposition of two spiral similarities is possible, the result is valid even for positions of point $P$ outside the quadrilateral and for a non-convex quadrilateral.

Bottema's theorem

Bottema ginma.png

Let triangle $\triangle SCD$ be given. Let triangles $\triangle SDA, \triangle SDA', \triangle SCB, \triangle SCB', \triangle CDP, \triangle CDP'$ be the isosceles rectangular triangles (see diagram).

Prove that $P$ and $P'$ be the midpoints of $AB$ and $A'B',$ respectively.

Proof

For given point $S$ one can find points $A(A')$ using rotation point $S$ around $D$ at the $90^\circ$ in counterclockwise (clockwise) direction. One can find point $B(B')$ using simmetry $A(A')$ with respect $P(P').$

We use Dao bisectors theorem for quadrilateral $ABCD$ with $\alpha = 90^\circ, \beta = 45^\circ$ and get existence given triangle $\triangle SCD$ with need properties.

Napoleon's theorem

Napoeon Dao.png
Napoleon Dao Inn.png

Let isosceles triangles with an angle of 120 degrees at the apex be constructed on the sides of an arbitrary triangle in the outer direction. The triangle with vertices at the apex those triangles names outer Napoleon triangle. Napoleon's theorem states that it is the equilateral triangle.

Let triangle $\triangle SCD$ be given. Let triangles $\triangle SDA, \triangle SCB, \triangle CDP$ be the isosceles triangles with angles $120^\circ$ (see diagram).

Prove that $\triangle ABP$ is the equilateral triangle.

Proof

For given point $S$ one can find points $A(B)$ using rotation point $S$ around $D(C)$ at the $30^\circ$ in counterclockwise (clockwise) direction and homothety with coefficient $\frac{1}{\sqrt{3}}.$

We use Dao bisectors theorem for quadrilateral $ABCD$ with $\alpha = 30^\circ, \beta = 120^\circ$ and get $AP = BP, \angle APB = 60^\circ.$

Note: Napoleon's theorem can also be proved for the inner triangle using the method of a pair of spiral symmetries (see diagram).

Finsler - Hadwiger theorem

Finsler theorem.png

Let two squares with common vertex be given. The theorem states that the quadrilateral of midpoints is (the Finsler–Hadwiger) square.

Let $AD$ and $BC$ be diagonals of given squares with common vertex $S.$ Let $M'$ and $M$ be the midpoints of $AB$ and $CD$ respectively, and let $O$ and $O'$ be the centers of these squares.

Prove that $OMO'M'$ is square.

Proof

Quadrilateral of midpoints is the parallelogram $OMO'M'.$

We use Dao bisectors theorem for quadrilateral $ABCD$ with $\alpha = \beta = 45^\circ$ and get \[\frac {M'M}{O'O} = \frac{\cot \alpha + \cot \beta}{2} = 1, M'M \perp OO' \implies\] $OMO'M'$ is the square.

Brahmagupta's theorem

Brachmagupta Dao.png

Let a cyclic quadrilateral $ABCD$ with $AC \perp BD$ be given, $S = AC \cap BD, M$ be the midpoint $CD.$

Prove that $MS \perp AB$ (Brahmagupta's theorem).

Proof

$\angle ADB = \angle ACB = \alpha, \angle DAC = \angle DBC = \beta$ $\beta = 90^\circ - \alpha, P$ is circumcenter $\odot ABCD,$ \[\angle PDM = 90^\circ - \beta = \alpha \implies Q = M.\]

We use Dao bisectors theorem for quadrilateral $ABCD$ with $\alpha + \beta = 90^\circ$ and get \[QS \perp AB \implies MS \perp AB.\]

Van Aubel's theorem

Van Aubel vectors.png
Van Aubel circle.png

Let squares $AFDF', AHBH', BECE', CGDG'$ constructed on the sides of a quadrilateral $ABCD$ be given.

Let $M, M'$ be the midpoints $AC$ and $BD,$ \[K = FE \cap H'G',K' = HG \cap E'F',\] \[L = EF \cap HG, L' = F'E' \cap H'G'.\] Notation is shown in the diagram.

1. Prove that $FE = GH, FE \perp GH,$ \[F'E' = G'H', F'E' \perp G'H'.\]

2. Prove that $MM'$ and $KK'$ are the perpendicular diameters of the circle $\omega$ centered at centroid $O$ of $ABCD$ and points $L$ and $L'$ lies on this circle.

Proof

Denote the centers of given squares $I, J, I', J',$ \[a =\vec {DI}, a' = \vec IF, b = \vec AI, b' = \vec JH,\] \[c = \vec BI', c' = \vec I'E, d = \vec CJ', d' = \vec J'G,\] $O -$ the centroid of $ABCD.$

It is evident that $\vec a + \vec b + \vec c + \vec d = 0.$

Vectors with sign ' are the rotations of the vectors without this sign in counterclockwise direction, so $|\vec a| = |\vec {a'}|,...$ and \[\vec {a'} + \vec b' + \vec c' + \vec d' = 0.\]

\[\vec {OA} + \vec {OB} + \vec {OC} + \vec {OD} = 0 \implies \vec {OA} = \vec a + \frac{\vec d - \vec b}{2},\] \[\vec {OB} = \vec b + \frac{\vec a - \vec c}{2},\vec {OC} = \vec c + \frac{\vec b - \vec d}{2}, \vec {OD} = \vec d + \frac{\vec c - \vec a}{2}.\]

1. $\vec {OF} = \vec {OD} + \vec a +\vec {a'} = \frac{\vec d - \vec b}{2} +\vec {a'}, \vec {OF'} = \vec {OF} - 2 \vec {a'} = \frac{\vec d - \vec b}{2} - \vec {a'}.$

\[\vec {OE} = \frac{\vec b - \vec d}{2} +\vec {c'}, \vec {OE'} = \frac{\vec b - \vec d}{2} - \vec {c'}.\]

$\vec {EF} = \vec {OF} - \vec {OE} = \vec d - \vec b +\vec {a'} - \vec {c'}.$

Similarly, $\vec {E'F'} = \vec d - \vec b +\vec {c'} - \vec {a'}, \vec {HG} = \vec c - \vec a +\vec {d'} - \vec {b'}, \vec {H'G'} = \vec a - \vec c +\vec {d'} - \vec {b'}.$

$\vec {a'}$ and $-\vec a$ are the results of rotation $\vec {a}$ and $\vec {a'}$ in counterclockwise direction at $90^\circ$ therefore $\vec {HG}$ is the result of rotation $\vec {EF}$ in counterclockwise direction at $90^\circ \implies$ \[EF = HG, EF \perp HG.\]

Similarly about $\vec {H'G'}$ and $\vec {E'F'}.$

2. $\vec OM = \frac{\vec {OA} + \vec {OC}}{2} = \frac{\vec a + \vec c}{2},$ \[\vec OM' = \frac{\vec {OB} + \vec {OD}}{2} = \frac{\vec b + \vec d}{2} = - \vec {OM},\] \[\vec {OK} = \vec {OF} + \mu \vec {EF} = \vec {OH' } + \nu \vec {G'H'}, \vec {OF} = \frac{\vec d - \vec b}{2} + \vec {a'}, \vec {OH' }= \frac{\vec a - \vec c}{2} - \vec {b'},\] \[\vec {OK} = \frac{\vec a' + \vec c'}{2}.\] Similarly, $\vec {OK'} = \frac{\vec b' + \vec d'}{2} = -\vec{OK} \implies OM = OK = OM' = OK', OM \perp OK.$ \[\angle KLK' = \angle ELH = 90^\circ \implies L \in \omega.\] \[\angle KL'K' = \angle E'L'H' = 90^\circ \implies L' \in \omega.\]

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