12/26/2024

$A,B,C$ are points on $\odot{O}$. $GA$ is tangent to $\odot{O}$. $C’$ is the reflection of $C$ along $GA$. $D=BC’ \cap \odot{O}$. $E,F$ are midpoints of $AC’, BC’$ respectively. Prove that $GA, EF, CD$ are concurrent.

Prove:

\(\begin{multline} \shoveleft \text{Suppose }EF \cap AG=G, \text{ we need prove }CDG \text{ is collinear}\\ \shoveleft \text{Extend }C'A \text{ to } H\text{ on the circle }\odot{ABC}\\ \shoveleft C'\text{is the reflection of }C\text{ along }AG \text{ and }AG \text{ is tangent to }\odot{ABC}\\ \shoveleft \implies \angle{CAG}=\angle{C'AG}=\angle{AHC}\implies AG \parallel CH\\ \shoveleft \implies \angle{CAG}=\angle{ACH}\implies AC=AC'=AH\\ \shoveleft \implies AF \parallel BH \implies \triangle{AEF}\sim\triangle{HAB}\\ \shoveleft \implies \angle{AFG}=\angle{HBA}=\angle{FAB}=\angle{ACH}=\angle{AHC}\\ \shoveleft \text{Since }\angle{GAD}=\angle{ABD}=\angle{GFD}\implies AGDF \text{ is cyclic }\\ \shoveleft \implies \angle{ADG}=\angle{AFG}=\angle{ACH}=\angle{AHC}\\ \shoveleft \implies CDG \text{ is collinear} \blacksquare \end{multline}\)

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12/30/2024

$\triangle{ABC}$ is an isosceles and $AC=BC$. $D, E$ are midpoints of $AB, BC$ respectively. The radius of circumcircle of $\triangle{ABC}$ is $1350$. The circumcircle of $\triangle{ACE}$ and $CD$ intersect at $F$. Find $CF$.

Solve:

\(\begin{multline} \shoveleft \text{AC=BC}, AD=BD \implies \angle{ACD}=\angle{BCD}\implies AF=EF\\ \shoveleft \text{And easy to know }AF=BF \implies EF=BF\\ \shoveleft \text{Make }H \text{ on }BE \text{ such that }FH\perp BE \implies \triangle{EFH}\cong\triangle{BFH}\\ \shoveleft \implies EH=HB\\ \shoveleft \text{Extend }AD \text{ and intersect with the circumcircle of }\triangle{ABC} \text{ at }G\\ \shoveleft \implies CO=GO=1350, \angle{CBG}=90^{\circ}\implies OE\parallel FH\parallel BG\\ \shoveleft \implies OF=FG=\dfrac{1350}{2}=675\implies CF=1350+675=\bbox[5px, border: 1px solid black]{2025} \end{multline}\)