Examples on Circumcircles Incircles and Excircles Set 4

Example - 67

Let H be the orthocenter in a triangle ABC as shown:

Prove that

(a)  \(HD = 2R\cos B\cos C,\;HE = 2R\cos C\cos A,\;HF = 2R\cos A\cos B\)

(b)  \(HA = 2R\cos A,\;HB = 2R\cos B,\;HC = 2R\cos C\)

Solution: (a) In  \(\Delta BHD\)  ,

\[\begin{align}&\;\;\;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\tan (\angle HBD) = \frac{{HD}}{{BD}} = \frac{{HD}}{{AB\cos B}} \\   &\Rightarrow  \qquad \tan \left( {\frac{\pi }{2} - C} \right) = \frac{{HD}}{{C\cos B}}  \\   &\Rightarrow  \qquad HD = c\cos B\cot C = \frac{c}{{\sin C}}\cos B\cos C = 2R\cos B\cos C  \\ 
\end{align} \]

Similarly, we can evaluate expressions for HE and HF

(b) In \(\Delta AEH\)  ,

\[\begin{align}&\;\;\;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cos (\angle HAE) = \frac{{AE}}{{HA}} = \frac{{AB\cos A}}{{HA}} \\  &\Rightarrow \qquad \cos \left( {\frac{\pi }{2} - C} \right) = \frac{{c\cos A}}{{HA}} \\  &\Rightarrow \qquad HA = \frac{{c\cos A}}{{\sin C}} = 2R\cos A \\ \end{align} \]

We can obtain analogous expressions for HB and HC.

Example - 68

Let G, H, O denote the centroid, orthocenter and circumcentre in \(\Delta ABC\) . Prove that \(OG:GH = 1:2\)

Solution:

Note that  \(OD = R\cos A\)  (how), and using the result of the previous example,  \(HA = 2R\cos A\)  . Thus,

\[HA = 2OD\]

Also, since G is the centroid,

\[AG = 2GD\]

Thus, \(\Delta AGH\) and  \(\Delta DGO\)  are similar with the sides ratio 2 : 1. This implies that  \(OG:GH = 1:2\) .

Note that we have not proved explicitly that O, G and H are collinear. If this is not evident to you, by all means try to prove it rigorously.

Example - 69

Let  \(I,\;{I_1},\;{I_2},\;{I_3}\)  be the incenter and the three ex-centres of  \(\Delta ABC\)  . Find

(a)  \({I_1}A\)       (b)  \({I_1}A\)

(c)  \(I\;{I_1}\)       (d)   \({I_1}\;{I_2}\)

Solution:

(a)  \({\text{In}}\;\Delta AID\)  ,

\[\begin{align}&\sin \frac{A}{2} = \frac{r}{{AI}}  \\\Rightarrow\quad&IA = r\;{\text{cosec}}\frac{A}{2} \\ \end{align} \]

(b)  \({\text{In}}\;\Delta A{I_1}D',\)

\[\begin{align}&\sin \frac{A}{2} = \frac{{{r_1}}}{{A{I_1}}} \\\Rightarrow\quad &{I_1}A = {r_1}\;{\text{cosec}}\frac{A}{2} \\\end{align} \]

(c)  \(\begin{align}I\,{I_1} = {I_1}A - IA =   {r_1}\,{\text{cosec}}\frac{A}{2} - r\;{\text{cosec}}\frac{A}{2}\end{align}\)

Since \(\begin{align}{r_1} = s\tan \frac{A}{2}\;{\text{and}}\;r = (s - a)\tan \frac{A}{2},\end{align}\) we have

\[I\,{I_1} = \frac{{s - (s - a)}}{{\cos \frac{A}{2}}} = a\sec \frac{A}{2}\]

(d)

\[\begin{align}&{I_1}\,{I_2} = {I_1}C + C{I_2} = {r_1}\sec \frac{C}{2} + {r_2}\sec \frac{C}{2} \\\\  \,\,\,\,\,\,\,\,\,\, &\qquad\qquad\qquad\quad\;\;= s\left( {\tan \frac{A}{2} + \tan \frac{B}{2}} \right)\sec \frac{C}{2} \\\\ \,\,\,\,\,\,\,\,\,\, &\qquad\qquad\qquad\quad\;\;= s\frac{{\sin \left( {\frac{{A + B}}{2}} \right)\sec \frac{C}{2}}}{{\cos \frac{A}{2}\cos \frac{B}{2}}}\\\\ \,\,\,\,\,\,\,\,\,\,&\qquad\qquad\qquad\quad\;\;= \frac{s}{{\sqrt {\frac{{s(s - a)}}{{bc}}} \sqrt {\frac{{s(s - b)}}{{ca}}} }} \\ \\ \,\,\,\,\,\,\,\,\,\, &\qquad\qquad\qquad\quad\;\;= c\;\;{\text{cosec}}\;\frac{C}{2}\\ \end{align} \]