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Let triangle $DBC$ inscribed in circle $(A)$,$H$ is the orthocenter of triangle $DBC$. Let $B'$ be the point of symmetry to $B$ through $DC$, $C'$ be the point of symmetry to $C$ through $DB$,$A'$ is the point of symmetry of $A $through $BC$. $B'C$ intersects $BC'$ at $I$. $HI$ intersects $BC$ at $L$. Prove that $D,L,A'$ is collinear.

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  • Prove that $C'B,B'C,AD$ are concurrent at $I$.

We can easily see the following:

$1.$ Inscribed quadrilateral $DC'BH$

$2.$ Inscribed quadrilateral $DB'CH$

$3.$ If , $C'B,B'C,AD$ is concurrent, then quadrilateral $ABIC$ is also a cyclic quadrilateral.

Thus, with the above$ 3 $things and by the angle addition method you can completely prove that $C'B,B'C,AD$ are concurrent at $I$ and quadrilateral $ABIC$ is also a cyclic quadrilateral.

-Prove that $D,L,A'$ is collinear.

All I need is just $HA$ parallel to $D'I$ . I need it because :

  • It is easy to see that there is $AA' $parallel to $DD'$, so we call$ AD'$ intersect $DA'$ at $L_1$, , according to $Talet's$ theorem we have:$ \frac {AL_1}{LD'} = \frac {AA'}{DD'}$ So if there is $HA$ parallel to$ D'I$, then we call $HS$ to cut $AD'$ at $L_2$ according to Talet's theorem, we also have: $ \frac {AL_2}{LD'} =\frac{AH}{D'I}= \frac{DH}{DD'}= \frac {AA'}{DD'}$

$\Rightarrow$ $L_1$ and $L_2$ are 2 points that coincide, so $D,L,A'$ is collinear.

But I have no idea how to prove $HA$ parallel to $D'I$ . I hope to get help from everyone . Thank you very much !

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    $\begingroup$ To be clear: $H$ is the intersection of $\overline{BB'}$ and $\overline{CC'}$ (aka, the orthocenter of $\triangle BCD$), correct? Also, by $HS$, you mean $HI$? And $D'$ is the reflection of $D$ in $\overline{BC}$? $\endgroup$ – Blue Jul 22 at 2:32
  • $\begingroup$ 1) Is A the center of the circle? 2) Is H the orthocenter? 3) What is S? Do you mean HI? 4) What is O? $\endgroup$ – Calvin Lin Jul 22 at 3:07
  • $\begingroup$ @CalvinLin I have re-edited my question. I am very sorry for my stupidity. Hope to get help from you. $\endgroup$ – abcccccc Jul 22 at 7:30
  • $\begingroup$ @BlueI have re-edited my question. I am very sorry for my stupidity. Hope to get help from you $\endgroup$ – abcccccc Jul 22 at 7:43
  • $\begingroup$ You can use barycentric coordinates. $\endgroup$ – mxian Jul 22 at 8:47

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