MathFun4Kids

Proof 1: Because $O$ is the center of the circle, $\triangle{OBC}$ is isosceles. and $\angle{OBC}=\angle{OCB}$. As $\angle{BOC}=2\angle{BAC}$, and $\angle{BOC} +\angle{OBC}+\angle{OCB}=180^\circ$, we have $$2\angle{BAC}+\angle{OBC}+\angle{OBC}=180^\circ$$ Therefore $\boxed{\angle{BAC}+\angle{OBC}=90^\circ}$

Proof 2: Because $O$ is the center of the circle, $\triangle{OAC}$ is isosceles. Therefore $$\angle{OAC}=\dfrac{1}{2}(180^\circ-\angle{AOC})$$ Because $\angle{AOC}=2\angle{ABC}$, we have $$\angle{OAC}=\dfrac{1}{2}(180^\circ-2\angle{ABC})=90^\circ-\angle{ABC}$$ Because $AP\perp BC$, $\triangle{APC}$ is right, $\angle{CAP}=90^\circ-\angle{ACB}$. Therefore $$\angle{OAP}=\angle{OAC}-\angle{CAP}=(90^\circ-\angle{ABC})-(90^\circ-\angle{ACB})=\boxed{\angle{ACB}-\angle{ABC}}$$

Proof 3: Let $ON\perp AP$ and $OA=OB=OC=r$.

Since $\triangle{AON}$ is right, and $\angle{AOP}=\angle{ACB}-\angle{ABC}\ge 30^\circ$, $$ON=sin(\angle{AOP})\cdot OA\ge sin(30^\circ)\cdot r=\dfrac{r}{2}$$ Since $MB=MC$, and $\triangle{BOC}$ is isosceles, $OM\perp BC$, $OMPN$ is a rectangle. Therefore, $MP=ON\ge \dfrac{r}{2}$.

Since $\triangle{ABC}$ is acute, $O$ is inside $\triangle{ABC}$, and $BC<2r$. Therefore $MB=MC=\dfrac{BC}{2}<r$. Because $MB=MP+CP$, $MP+CP<r$.

Because $MP\ge\dfrac{r}{2}$, $CP<r-MP\le r-\dfrac{r}{2}=\dfrac{r}{2}$. Since $MP\ge\dfrac{r}{2}$, $CP<\dfrac{r}{2}$, $\boxed{MP>CP}$

Proof 4: Since $\triangle{MOP}$ is right, and $OM\perp BC$, $OP>MP>CP$. Therefore $\angle{POC}<\angle{OCB}$.

Since $\angle{BAC}+\angle{OCB}=\angle{BAC}+\angle{OBC}=90^\circ$, Therefore

$$\boxed{\angle{BAC}+\angle{POC}<\angle{BAC}+\angle{OCB}=90^\circ}$$

Proof: Mark various tangent points as the above. We have $JK=ML$. $$\because JK=KE+JE=EN+EO=EN+EN+NO=2EN+NO$$ and $$ ML=LD+MD=DO+DN=DO+DO+NO=2DO+NO$$ Therefore $EN=DO$, $KE=DL$, and $JE=MD$.

$$\because AE=AN-EN=AG-KE=AC-CG-DL=AC-CM-DL$$ and$$AE=AO-EO=AF-JE=AB-BF-MD=AB-BL-MD$$

We have $$2AE=AC-CM-DL+AB-BL-MD$$ $$=AB+AC-(CM+MD+DL+BL)=AB+AC-BC$$ Therefore $$AE=\dfrac{AB+AC-BC}{2}$$

which implies that the length of $AE$ is a constant for $\triangle{ABC}$, and $E$ is on an arc of a circle, as shown below, where $P$ and $Q$ are tangent points of the in-circle of $\triangle{ABC}$ on $AC$ and $AB$ respectively.