Online Course Discussion Forum
Centers of Triangle Example 3
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)
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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