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Centers of Triangle Example 3

 
 
SongKevin的头像
Centers of Triangle Example 3
SongKevin - 2023年11月18日 Saturday 07:59
 

Hello, in the Example 3 on the Centers of a triangle worksheet we went over in class, I understand that MKBS and MLBR are cyclic, but I for the next step when we used power of a point for \( RI^2-r^2=RK \cdot RM = RB \cdot RS \), why are the ratio of the sides equal? Did we prove that \( KB//MS \)? (This is ~55:00 in the recording)

 
LensmireJohn的头像
Re: Centers of Triangle Example 3
LensmireJohn - 2023年11月18日 Saturday 14:45
 

Hi Kevin,

The equalities both follow from Power of a Point. The first, $RI^2 - r^2 = RK\cdot RM$, is from the circle with center $I$. Then, consider the circle containing $MKBS$. Power of a Point here gives $RK\cdot RM = RB\cdot RS$.

Note: this same type of reasoning is used again (the second time using that $MLBR$ is cyclic.

Hope this helps!

 
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SongKevin的头像
Re: Centers of Triangle Example 3
SongKevin - 2023年11月19日 Sunday 09:26
 

I am also confused on Example 5 (~1:27:40) and how we got \( \frac{DM \cdot \sin (\angle PDM)}{DI \cdot \sin (\angle PDI)} = \frac{2R \cdot \sin^2 (\alpha)}{2R \cdot \sin (\alpha)} \cdot \frac{\sin(90- \alpha + \alpha - \beta)}{\sin(C + \alpha - \beta)} \) 

 
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