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)
\[\begin{align}&{I_1}\,{I_2} = {I_1}C + C{I_2} = {r_1}\sec \frac{C}{2} + {r_2}\sec \frac{C}{2} \\\\ \,\,\,\,\,\,\,\,\,\, &\qquad\qquad\qquad\quad\;\;= s\left( {\tan \frac{A}{2} + \tan \frac{B}{2}} \right)\sec \frac{C}{2} \\\\ \,\,\,\,\,\,\,\,\,\, &\qquad\qquad\qquad\quad\;\;= s\frac{{\sin \left( {\frac{{A + B}}{2}} \right)\sec \frac{C}{2}}}{{\cos \frac{A}{2}\cos \frac{B}{2}}}\\\\ \,\,\,\,\,\,\,\,\,\,&\qquad\qquad\qquad\quad\;\;= \frac{s}{{\sqrt {\frac{{s(s - a)}}{{bc}}} \sqrt {\frac{{s(s - b)}}{{ca}}} }} \\ \\ \,\,\,\,\,\,\,\,\,\, &\qquad\qquad\qquad\quad\;\;= c\;\;{\text{cosec}}\;\frac{C}{2}\\ \end{align} \]
- Live one on one classroom and doubt clearing
- Practice worksheets in and after class for conceptual clarity
- Personalized curriculum to keep up with school