Properties and Solution of Triangles
12.0 Distances of special points from vertices and sides of a triangle
12.0 Distances of special points from vertices and sides of a triangle
Consider a triangle as shown in the figure.
$(i)$ Circumcentre($O$): $$OA = R;{O_a} = R\cos A$$
$(ii)$ Incentre($I$): $$IA = r\,{\text{cosec}}\frac{A}{2};{I_a} = r$$
$(iii)$ Excentre(${I_1}$): $${I_1}A = {r_1}\,{\text{cosec}}\frac{A}{2};{I_{1a}} = {r_1}$$
$(iv)$ Orthocentre($H$): $$HA = 2R\cos A;{H_a} = 2R\cos B\cos C$$
$(v)$ Centroid($G$): $$GA = \frac{1}{3}\sqrt {2{b^2} + 2{c^2} - {a^2}} ;{G_a} = \frac{{2\Delta }}{{3a}}$$
Note: $XA$ is the distance of each point from the vertex $A$ of the triangle and ${X_a}$ is the distance of each point from the side $A$ of the triangle.
Question 27. If $x$, $y$ and $z$ are respectively the distances of the vertices of the triangle $ABC$ from its orthocentre, then prove the following:
(i) $\frac{a}{x} + \frac{b}{y} + \frac{c}{z} = \frac{{abc}}{{xyz}}$
(ii) $x + y + z = 2\left( {R{\text{ }} + {\text{ }}r} \right)$
Solution: $(i)$ As we know that $$a = 2R\sin A,b = 2R\sin B,c = 2R\sin C$$
$$x = 2R\cos A,b = 2R\cos B,c = 2R\cos C$$
$$\begin{equation} \begin{aligned} \frac{a}{x} + \frac{b}{y} + \frac{c}{z} = \tan A + \tan B + \tan C...(i) \\ \frac{{abc}}{{xyz}} = \tan A\tan B\tan C...(ii) \\\end{aligned} \end{equation} $$
We know that in triangle $ABC$,$$\sum \tan A = \prod \tan A$$
From equations $(i)$ and $(ii)$,we can say that $$\frac{a}{x} + \frac{b}{y} + \frac{c}{z} = \frac{{abc}}{{xyz}}$$
$(ii)$ $$x + y + z = 2R(\cos A + \cos B + \cos C)$$
We know that in a triangle $ABC$ $$\cos A + \cos B + \cos C = 1 + 4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$$
so we get,
$$\begin{equation} \begin{aligned} x + y + z = 2R\left( {1 + 4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}} \right) \\ = 2\left( {R + 4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}} \right) \\\end{aligned} \end{equation} $$
we know that $$r = 4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$$
therefore $$x + y + z = 2(R + r)$$
Question 28. If $r$ be the incentre of triangle $ABC$ then prove that $$IA \cdot IB \cdot IC = abc\tan \frac{A}{2}\tan \frac{B}{2}\tan \frac{C}{2}$$
Solution: Consider LHS
Using $$\begin{equation} \begin{aligned} IA = r\cos ec\frac{A}{2} \\ IB = r\cos ec\frac{B}{2} \\ IC = r\cos ec\frac{C}{2} \\\end{aligned} \end{equation} $$
we can write $$\begin{equation} \begin{aligned} IA \cdot IB \cdot IC = r\cos ec\frac{A}{2} \times r\cos ec\frac{B}{2} \times r\cos ec\frac{C}{2} \\ = \frac{{{r^3}}}{{\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}}} \\\end{aligned} \end{equation} ...(i) $$
we know that $$\begin{equation} \begin{aligned} r = 4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2} \\ \sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2} = \frac{r}{{4R}} \\\end{aligned} \end{equation} ...(ii) $$
so substituting $(ii)$ in $(i)$, we get $$\begin{equation} \begin{aligned} = \frac{{{r^3}}}{{\frac{r}{{4R}}}} \\ = \frac{{{r^3} \times 4R}}{r} \\\end{aligned} \end{equation} $$
Substituting the $r$ in the denominator as $$r = \frac{\Delta }{s}$$
and $$R = \frac{{abc}}{{4\Delta }}$$
and $r$ in the numerator as $$\begin{equation} \begin{aligned} r = (s - a)\tan \frac{A}{2} \\ r = (s - b)\tan \frac{B}{2} \\ r = (s - c)\tan \frac{C}{2} \\\end{aligned} \end{equation} $$
We get $$ = \frac{{4(abc)s(s - a)\tan \frac{A}{2}(s - b)\tan \frac{B}{2}(s - c)\tan \frac{C}{2}}}{{4{\Delta ^2}}}$$
and substituting $$\Delta = \sqrt {s(s - a)(s - b)(s - c)} $$
we get $$abc\tan \frac{A}{2}\tan \frac{B}{2}\tan \frac{C}{2}$$
