# Finding the orthocentre of a trinagle.

Now, I know this has been asked here but my question is something else so please bear with me.

Question:- If the vertices of a triangle are represented by $z_1, z_2, z_3$ respectively then show that the orthocentre of the $\triangle ABC$ is

$$\dfrac{(a \sec A)z_1+(b \sec B)z_2+(c \sec C)z_3}{a \sec A+b \sec B +c \sec C }$$ $$OR$$ $$\dfrac{z_1\tan A+z_2\tan B+z_3\tan C}{\tan A+\tan B+\tan C}$$

Attempt at a solution:-

Now, from simple trigonometry, to be more specific, from projection law, we can see that

$$\frac{BL}{LC}=\dfrac{c\cos B}{b\cos C}=\dfrac{c\sec C}{b \sec B}$$

So we get from the section formula the coordinates of $L$ as follows

$$L \equiv \left(\dfrac{b\sec B\cdot z_2+c\sec C\cdot z_3}{b\sec B+c\sec C}\right)$$

Now, I don't know much about properties of triangles and found out the following method in a book for finding the ratio in which $H$ divides $AL$

\begin{aligned} \dfrac{AH}{HL} &=\dfrac{ar(\triangle ABH)}{ar(\triangle HBL)} \\ &=\dfrac{\dfrac{1}{2}AB.BH\sin{\alpha}}{\dfrac{1}{2}BL.BH \sin{\beta}} \\ &=\dfrac{c \cos A}{c\cos B\cos C}=\dfrac{a \cos A}{a\cos B\cos C} \\ &=\dfrac{b\cos C +c\cos B}{a\sec A}\cdot\dfrac{1}{\cos B\cos C} \\ &=\dfrac{b \sec B+c\sec C}{a\sec A} \end{aligned}

Now I don't understand what was used in the derivation of the required ratio above, is it similarity of triangles but i don't think the triangles are similar.

P.S.:- Corrected the link to the question mentioned at the top.

• The link in the first sentence does not seem to work. – mathlove Jul 15 '16 at 7:18
• @mathlove, sorry for the late reply. Just corrected it. – user350331 Jul 15 '16 at 7:39
• So, is your question about how to get $AH/HL=(b\sec B+c\sec C)/(a\sec A)$? – mathlove Jul 15 '16 at 7:58
• Yeah, that's what i wanted and came upon this proof which relates the area of a triangle to its sides which i didn't get how the book did it, so if you could suggest me how it was done or maybe suggest a better method or just tell me what the answer in the top link did. But, my first priority would be to understand how the book did it. – user350331 Jul 15 '16 at 8:05

$$\dfrac{AH}{HL}=\dfrac{ar(\triangle ABH)}{ar(\triangle HBL)}$$

Because the two triangles have the common height $BL$.

$$\dfrac{ar(\triangle ABH)}{ar(\triangle HBL)}=\dfrac{\dfrac{1}{2}AB.BH\sin{\alpha}}{\dfrac{1}{2}BL.BH \sin{\beta}}$$

$$\dfrac{ar(\triangle ABH)}{ar(\triangle HBL)}=\dfrac{\dfrac 12 AB\cdot NH}{\dfrac 12BL\cdot HL}=\dfrac{\dfrac{1}{2}AB\cdot BH\sin{\alpha}}{\dfrac{1}{2}BL\cdot BH \sin{\beta}}$$

$$\dfrac{\dfrac{1}{2}AB.BH\sin{\alpha}}{\dfrac{1}{2}BL.BH \sin{\beta}}=\dfrac{c \cos A}{c\cos B\cos C}$$

See $\triangle{ABM}$ and $\triangle{BMC}$ : $$\dfrac{\dfrac{1}{2}AB.BH\sin{\alpha}}{\dfrac{1}{2}BL.BH \sin{\beta}}=\frac{AB\sin\alpha}{\color{red}{BL}\sin\beta}=\frac{c\sin(90^\circ-A)}{\color{red}{c\cos B}\sin(90^\circ-C)}=\frac{c\cos A}{c\cos B\cos C}$$

$$\dfrac{c \cos A}{c\cos B\cos C}=\dfrac{a \cos A}{a\cos B\cos C}$$

Because $\frac cc=\frac aa=1$.

$$\dfrac{a \cos A}{a\cos B\cos C}=\dfrac{b\cos C +c\cos B}{a\sec A}\cdot\dfrac{1}{\cos B\cos C}$$

$$\dfrac{a \cos A}{a\cos B\cos C}=\dfrac{\color{red}{a}}{a\sec A}\cdot \dfrac{1}{\cos B\cos C}=\dfrac{\color{red}{CL+BL}}{a\sec A}\cdot \dfrac{1}{\cos B\cos C}$$ Now $CL=b\cos C$ and $BL=c\cos B$.

$$\dfrac{b\cos C +c\cos B}{a\sec A}\cdot\dfrac{1}{\cos B\cos C}=\dfrac{b \sec B+c\sec C}{a\sec A}$$

$$\dfrac{b\cos C +c\cos B}{a\sec A}\cdot\dfrac{1}{\cos B\cos C}=\frac{\frac{b\cos C}{\cos B\cos C}+\frac{c\cos B}{\cos B\cos C}}{a\sec A}=\dfrac{b \sec B+c\sec C}{a\sec A}$$

• What the...!!! I just refreshed the page and there was the answer in front of me, I cannot be more thankful to you, you just get what i want to ask even if I am not much clear to state my query. Thanks so much. – user350331 Jul 15 '16 at 8:12
• @user350331: You are welcome:) – mathlove Jul 15 '16 at 8:14