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Difference between revisions of "Dao Thanh Oai geometric results"

(Created page with "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 discove...")
 
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<i><b>Proof</b></i>
 
<i><b>Proof</b></i>
  
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1. <math>\angle SAD = \angle QPD, \angle SDA = \angle QDP \implies \triangle DAS \sim \triangle DPQ.</math>
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1. <math>\angle SAD = \angle QPD, \angle SDA = \angle QDP \implies</math>
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<cmath>\triangle DAS \sim \triangle DPQ.</cmath>
  
 
The spiral similarity taking <math>A</math> to <math>S</math> and <math>P</math> to <math>Q</math> has center <math>D</math> and angle <math>\alpha.</math> Therefore spiral similarity taking <math>AP</math> to <math>SQ</math> and <math>P</math> to <math>Q</math> has the same center <math>D</math> and angle <math>\alpha.</math>  
 
The spiral similarity taking <math>A</math> to <math>S</math> and <math>P</math> to <math>Q</math> has center <math>D</math> and angle <math>\alpha.</math> Therefore spiral similarity taking <math>AP</math> to <math>SQ</math> and <math>P</math> to <math>Q</math> has the same center <math>D</math> and angle <math>\alpha.</math>  

Revision as of 05:43, 29 August 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

Quadrilateral with side bisectors

Dao ginma1.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 $\triangle BSC \sim \triangle ASD.$

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 $k, B = t(S), \frac {BC}{BS} = k.$

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

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

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