# 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? 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. Jul 22 at 7:30
• @BlueI have re-edited my question. I am very sorry for my stupidity. Hope to get help from you Jul 22 at 7:43
• You can use barycentric coordinates. Jul 22 at 8:47 We decided to divide our proof into four parts for the sake of easy explaining. In the first part we prove that the three points $$D$$, $$A$$, and $$I$$ are collinear. In the second part, we show that the three points $$M$$, $$D$$, and $$C’$$ are collinear. The third part contains the proof of collinearity of yet another set of three points $$D’$$, $$L$$, and $$A$$, where $$D’$$ is the reflection of $$D$$ in $$BC$$. With the help of this, we deduce in the fourth part the collinearity of the three points $$D$$, $$L$$, and $$A’$$.

$$\underline{\bf\text{Part 1}}:\enspace \text{Proof of the collinearity of}\space D,\space A,\space \text{and}\space I$$

As illustrated in $$\mathrm{Fig.\space 1}$$, we join $$DA$$, $$AI$$, $$AB$$, $$AC$$. Consider the three vertex angles of $$\triangle CBD$$. For brevity, let $$\measuredangle DCB = \omega$$ and $$\measuredangle CBD = \phi$$. Hence, the remaining angle $$BDC$$ can be expressed as, $$\measuredangle BDC = 180^o-\phi-\omega. \tag{1}$$

Since $$C’$$ is the reflection of $$C$$ in side $$BD$$, $$\triangle C”BC$$ is an isosceles triangle. Therefore, we have, $$\measuredangle DBC = \measuredangle CBD = \phi \quad\rightarrow\quad \measuredangle IBC = 180^o – 2\phi. \tag{2}$$

Similarly, $$\triangle CBB’$$ is an isosceles triangle because $$B’$$ is the reflection of $$B$$ in side $$DC$$. This makes, $$\measuredangle B’CD = \measuredangle DCB = \omega \quad\rightarrow\quad \measuredangle BCI = 180^o – 2\omega. \tag{3}$$

Since (2) and (3) give two of the angles of $$\triangle BCI$$, we can find its remaining angle as $$\measuredangle CIB = 2\phi+2\omega – 180^o. \tag{4}$$

The angles $$BDC$$ and $$BAC$$ are the angles subtended by the chord $$BC$$ at the circumference and the center $$A$$ of the circumcircle of $$\triangle DBC$$ respectively.The angle $$BDC$$ is given in (1). Therefore, we can express angle $$BAC$$ as, $$\measuredangle BAC = 2\times \measuredangle BDC = 360^o -2\phi -2\omega. \tag{5}$$

From (4) and (5), it is obvious that the two angles $$BAC$$ and $$CIB$$ of the quadrilateral $$BACI$$ are supplement of each other. Hence, $$BACI$$ is cyclic. We shall write, $$\measuredangle BAI = \measuredangle BCI = 180^o - 2\omega. \tag{6}$$

Since the angle $$DAB$$ is the angle subtended by the chord $$BD$$ at the center $$A$$ of the circumcircle of $$\triangle DBC$$, we have, $$\measuredangle DAB = 2\times \measuredangle DCB = 2\omega. \tag{7}$$

According to (6) and (7), the two angles $$BAI$$ and $$DAB$$ are supplement of each other, which is so because points $$D$$, $$A$$, $$I$$ are collinear. $$\underline{\bf\text{Part 2}}:\enspace \text{Proof of the collinearity of}\space M,\space D,\space \text{and}\space C’$$

As shown in $$\mathrm{Fig.\space 2}$$, we introduced a new point $$D’$$ to our configuration as the reflection of $$D$$ in $$BC$$. After joining $$D’C$$ and $$BA$$, we extend both of them to meet at $$M$$. Furthermore, it is necessary to join $$D’E$$, $$MC’$$ and $$DC’$$. Please note that the angles shown in red have been either assumed or determined in Part 1.

In the right angle triangle $$CFD$$, the angle $$DCF$$ can be expressed as, $$\measuredangle DCF = 90^o – \measuredangle FDC = \phi + \omega – 90^o. \tag{8}$$ Since, $$C’$$ is the reflection of $$C$$ in side $$BD$$, $$\triangle C”DC$$ is an isosceles triangle. Hence, we have, $$\measuredangle CC’D = \measuredangle DCC’ = \phi+\omega + 90. \tag{9}$$

As mentioned above, we introduced $$D’$$ as the reflection of $$D$$ in side $$BD$$. This makes $$\triangle C”DC$$ is an isosceles triangle, of which $$CE$$ is the $$C$$-altitude. Therefore, $$\measuredangle ECD’ = \measuredangle DCE = \omega \space\rightarrow\space \measuredangle DCD’ = \measuredangle ECD’ + \measuredangle DCE = 2\omega \space\rightarrow\space \\ \measuredangle MCD = 180^o - \measuredangle DCD’ = 180^o – 2\omega. \tag{10}$$

We can determine angle MCC’ using (9) and (10). $$\measuredangle MCC’ = \measuredangle MCD + \measuredangle DCC' = 90^o + \phi - \omega \tag{11}$$

Since $$AD$$ and $$AB$$ are radii of the circumcircle of $$\triangle DCB$$, the triangle $$DAB$$ is an isosceles triangle. Hence, $$\measuredangle MBD = 90^o - \omega. \tag{12}$$

Now, using (12), we can express the angles $$MBC’$$ and $$CBM$$ as shown below. $$\measuredangle MBC’ = \measuredangle MBD + \measuredangle DBC’ = 90^o + \phi - \omega \tag{13}$$

$$\measuredangle CBM = \measuredangle CBD - \measuredangle MBD’ = \phi + \omega – 90^o \tag{14}$$

According to (11) and (13), $$\measuredangle MBC’ = \measuredangle MCC’$$. Therefore, the quadrilateral $$MCBC’$$ is cyclic and, using (14), we shall write, $$\measuredangle CC’M = \measuredangle CBM = \phi + \omega – 90^o. \tag{15}$$

When we compared (9) and (15), we see that $$\measuredangle CC’M = \measuredangle CC’D,$$ which is possible if and only if points $$M$$, $$D$$, and $$C’$$ are collinear. $$\underline{\bf\text{Part 3}}:\enspace \text{Proof of the collinearity of}\space D’,\space L,\space \text{and}\space A$$

As shown in $$\mathrm{Fig.\space 3}$$, line segments $$D’L$$ and $$LA$$ were added to the diagram.

Now, consider the two triangles $$MCB$$, which is green, and $$DHI$$, which is red. By grouping together one vertex from each triangle, we form three different pairs of vertices, which we call pairs of corresponding vertices. If we can form these pairs of corresponding vertices in such a manner that the three lines joining the two vertices in each pair concur at a point, which is called the perspective center, we say that the two triangles are perspective from a point. Consider the following pairs of vertices formed from $$\triangle MCB$$ and $$\triangle DHI$$. $$\{ M,\enspace D\},\qquad\{C,\enspace H\},\qquad \{B,\enspace I\}\qquad$$

We proved in Part 2 that the line $$MD$$ goes through the point $$C’$$. Meeting of the lines $$CH$$ and $$IB$$ at $$C’$$ is specific to the given configuration. Therefore, these two triangles are perspective from a point.

According to the Desargues’ theorem, if two triangles are perspective from a point, they are perspective from a line called perspective axis as well. This means that the three points of intersection of pairs of corresponding sides lie on a straight line. The three pairs of corresponding sides of the two triangles $$MCB$$ and $$DHI$$ and their individual points of intersection are given below. $$\{MC,\enspace DH,\enspace D’\},\qquad\{CB,\enspace HI,\enspace L\},\qquad \{BM,\enspace ID,\enspace A\}\qquad$$

Therefore, the points $$D’$$, $$L$$, and $$A$$ is on the perspective axis of $$\triangle MCB$$ and $$\triangle DHI$$. This means that these three points are collinear.

$$\underline{\bf\text{Part 4}}:\enspace \text{Proof of the collinearity of}\space D,\space L,\space \text{and}\space A’$$

Consider the two line segments $$D’A$$ and $$DA’$$ shown in $$\mathrm{Fig.\space 3}$$. Each one of these two segments is the reflection of the other about $$BC$$. It is always possible to find a point on one segment, which is the reflection in $$BC$$ of a point lying on the other segment. In Part 3, we showed that point $$L$$, which lies on $$BC$$, lies also on $$D’A$$. Now, imagine a point $$L’$$ on $$DA’$$, which is the reflection of $$L$$ in $$BC$$. Since $$L$$ is located on $$BC$$, its reflection in $$BC$$, namely $$L’$$, is the selfsame. Therefore, $$L$$ lies on $$DA’$$ allowing us to state that the three points $$D$$, $$L$$, and $$A’$$ are collinear.