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Dao Thanh Oai geometric results

Revision as of 15:12, 30 August 2025 by Vvsss (talk | contribs) (→‎Dao bisectors theorem)

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.

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