SATHEE: Properties of Triangles 3 Question 18

Properties of Triangles 3 Question 18

18. Let $A B C$ be a triangle having $O$ and $I$ as its circumcentre and incentre, respectively. If $R$ and $r$ are the circumradius and the inradius respectively, then prove that $(I O)^{2} = R^{2} - 2 R r$ . Further show that the $△ B I O$ is a right angled triangle if and only if $b$ is the arithmetic mean of $a$ and $c$ .

(1999, 10M)

Show Answer

Solution:

It is clear from the figure that, $O A = R$

$A I = \frac{I F}{\sin (A / 2)}$

$∵ △ A I F$ is right angled triangle, so $= \frac{r}{\sin (A / 2)}$

$\begin{aligned} But & r & = 4 R \sin (A / 2) \sin (B / 2) \sin (C / 2) \\ ∴ & A I & = 4 R \sin (B / 2) \sin (C / 2) \\ Again, & ∠ G O A & = B \Rightarrow O A G = 90^{\circ} - B \\ Therefore, ∠ I A O & = ∠ I A C - ∠ O A C \\ = A / 2 - (90^{\circ} - B) = \frac{1}{2} (A + 2 B - 180^{\circ}) \\ = \frac{1}{2} (A + 2 B - A - B - C) = \frac{1}{2} (B - C) \end{aligned}$

In $△ O A I, O I^{2} = O A^{2} + A I^{2} - 2 (O A) (A I) \cos (∠ I A O)$

$\begin{aligned} = R^{2} + [4 R \sin (B / 2) \sin (C / 2)]^{2} \\ - 2 R \cdot [4 R \sin (B / 2) \sin (C / 2)] \cos \frac{B - C}{2} \\ = [R^{2} + 16 R^{2} \sin^{2} (B / 2) \sin^{2} (C / 2) \\ - 8 R^{2} \sin (B / 2) \sin (C / 2) \cos \frac{B - C}{2} \\ = R^{2} [1 + 16 \sin^{2} (B / 2) \sin^{2} (C / 2) \\ - 8 \sin (B / 2) \sin (C / 2) \cos \frac{B - C}{2} \\ = R^{2} [1 + 8 \sin (B / 2) \sin (C / 2) \\ 2 \sin (B / 2) \sin (C / 2) - \cos \frac{B - C}{2} \end{aligned}$

$\begin{aligned} = R^{2} [1 + 8 \sin (B / 2) \sin (C / 2) \\ \cos \frac{B - C}{2} + \cos \frac{B + C}{2} - \cos \frac{B - C}{2} \\ = R^{2} 1 - 8 \sin (B / 2) \sin (C / 2) \cos \frac{B + C}{2} \\ = R^{2} 1 - 8 \sin (B / 2) \sin (C / 2) \cos \frac{π}{2} - \frac{A}{2} \\ ∵ \frac{A}{2} + \frac{B}{2} + \frac{C}{2} = \frac{π}{2} \\ = R^{2} [1 - 8 \sin (A / 2) \sin (B / 2) \sin (C / 2)] \\ = R^{2} 1 - 8 \frac{r}{4 R} = R^{2} - 2 R r \end{aligned}$

Now, in right angled $△ B I O$ ,

$\begin{aligned} O B^{2} = B I^{2} + I O^{2} \\ \Rightarrow R^{2} = B I^{2} + R^{2} - 2 R r \\ \Rightarrow 2 R r = B I^{2} \\ \Rightarrow 2 R r = r^{2} / \sin^{2} (B / 2) \\ \Rightarrow 2 R = r / \sin^{2} (B / 2) \\ \Rightarrow 2 R \sin^{2} B / 2 = r \\ \Rightarrow R (1 - \cos B) = r \\ \Rightarrow \frac{a b c}{4 Δ} (1 - \cos B) = \frac{Δ}{s} \\ \Rightarrow a b c (1 - \cos B) = \frac{4 Δ^{2}}{s} \\ \Rightarrow a b c 1 - \frac{a^{2} + c^{2} - b^{2}}{2 a c} = \frac{4 Δ^{2}}{s} \\ \Rightarrow a b c \frac{2 a c - a^{2} - c^{2} + b^{2}}{2 a c} = \frac{4 Δ^{2}}{s} \end{aligned}$

$\begin{array}{cc} \Rightarrow & b [b^{2} - (a - c)^{2}] = \frac{4 Δ^{2}}{s} \\ \Rightarrow & b [b^{2} - (a - c)^{2}] = 8 (s - a) (s - b) (s - c) \\ \Rightarrow & b [b - (a - c) b + (a - c)] \\ \Rightarrow & = 8 (s - a) (s - b) (s - c) \\ \Rightarrow & b [(b + c - a) (b + a - c)] = 8 (s - a) (s - b) (s - c) \\ \Rightarrow & b [(2 s - 2 a) (2 s - 2 c)] = 8 (s - a) (s - b) (s - c) \\ \Rightarrow & b [2 \cdot 2 (s - a) (s - c)] = 8 (s - a) (s - b) (s - c) \\ \Rightarrow & b = 2 s - 2 b \\ \Rightarrow & 2 b = a + c \end{array}$

which shows that $b$ is arithmetic mean between $a$ and $c$ .

Properties of Triangles 3 Question 18

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Properties of Triangles 3 Question 18

18. Let ABC be a triangle having O and I as its circumcentre and incentre, respectively. If R and r are the circumradius and the inradius respectively, then prove that (IO)2=R2−2Rr. Further show that the △BIO is a right angled triangle if and only if b is the arithmetic mean of a and c.

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu