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

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.

• 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 !

• 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 $\overline{BC}$? – Blue Jul 22 at 2:32
• 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? – Calvin Lin Jul 22 at 3:07
• @CalvinLin I have re-edited my question. I am very sorry for my stupidity. Hope to get help from you. – abcccccc Jul 22 at 7:30
• @BlueI have re-edited my question. I am very sorry for my stupidity. Hope to get help from you – abcccccc Jul 22 at 7:43
• You can use barycentric coordinates. – mxian Jul 22 at 8:47