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Dao Thanh Oai geometric results

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 \[QS \perp AB \implies MS \perp AB.\]

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