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Point $P$ is outside circle $\odot{O}$, and $A,B$ are tagent points from $P$ to $\odot{O}$. $C,D$ are two points on $\odot{O}$ that are colinear with $P$. Then we have $\dfrac{AC}{AD}=\dfrac{BC}{BD}$. I wonder if there is a named theorem for this. It can be easily proved as following:

$$ \begin{multline}\nonumber \shoveleft \triangle{PAC}\sim\triangle{PDA}\implies \dfrac{AC}{AD}=\dfrac{PA}{PD}\\ \shoveleft \triangle{PBC}\sim\triangle{PDB}\implies \dfrac{BC}{BD}=\dfrac{PB}{PD}=\dfrac{PA}{PD}\\ \shoveleft \implies \dfrac{AC}{AD}=\dfrac{BC}{BD} \end{multline} $$

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r ne
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1 Answers1

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This is in fact called a harmonic quadrilateral. It is a topic related to projective geometry and cross ratios.

A cyclic quadrilateral ABCD is called harmonic if AB/BC=AD/DC. Harmonic quadrilaterals appear a lot in geometry especially in competition wise problems.

There are a lot of interesting properties of harmonic quadrilateral based on your diagram, maybe you can try to prove a few of them.

1)Consider the midpoint of AB, call M. DC is in fact the reflection of DM over the angle bisector of BDA. (DC is called the D symmedian of triangle DBA)

2)Let DC intersect BA at Q. We have: (BQ/QA)=(DB/DA)^2

3)The circumcircle of triangle PBA, passes through the midpoint of DC

4)Let M be the midpoint of AB, AB is in fact the angle bisector of angle DMC

5)Let DC intersect BA at Q. We have (PC/CQ)/(PD/DQ)=1. This is actually called a cross ratio and in projective geometry it is denoted as (P,Q;C,D)=-1 and we call P,Q,C,D a harmonic bundle.

6)Draw the tangents to the circle at C,D, and let T be the intersection of the two tangents. In fact, T,A,B are collinear

To find out more, the keywords are Cross Ratios, Harmonic Bundles, Harmonic Quadrilaterals. I have linked a source below if you are interested to find out more about this.

https://alexanderrem.weebly.com/uploads/7/2/5/6/72566533/projectivegeometry.pdf

avan
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  • Thanks for the sharing and it's great. However, these are properties of harmonic quadrilateral, after it is proved to have AB/BC=AD/DC. What I want to know is whether there is some named theorem to prove ABCD is harmonic... – r ne Nov 17 '23 at 09:44
  • There is no current named theorem that I know of to prove ABCD is harmonic, but maybe a different answer is that some of the properties above are in fact iff. For example, If ABCD is cyclic and AC is the A symmedian of triangle ABD, then it is necessary that ABCD is harmonic, I think quite some of them are discussed in the link I sent. – avan Nov 17 '23 at 14:44
  • Thanks! I cannot find such a theorem either so far... – r ne Nov 18 '23 at 19:15