Name: Theorem on Perpendicular Bisectors of a Triangle
Uploaded: 2019-07-19
Duration: 4 min 55 s
Description: What are the perpendicular bisectors of the sides of a triangle? Are the perpendicular bisectors of a triangle concurrent? To know the answers to all these questions, watch this video.

Question

If a circle be drawn touching the inscribed and circumscribed circle of a triangle and the side

B C

, externally, prove that its radius is

\frac{∆}{a} \tan^{2} \frac{A}{2}

.

Accepted Answer

Let $\frac{∆}{a} \tan^{2} \frac{A}{2}$ be the triangle whose in-centre and circum-centre are $I$ and $O$ .

Now, according to the figure, $K$ is the centre of the required circle, which touches in-circle, the circum-circle and the side $B C$ .

Certainly the line $I D K$ is perpendicular on side $B C$ at $D$ . Draw $O N M$ , parallel to $I D K$ , which cuts side $B C$ at $N$ .

Draw perpendicular $K M$ on $O N M$ .

Let the required radius be $D K = x$ .

In triangle $∆ O C N$ , $∠ N O C = A$ .

$∴ \sin A = \frac{C N}{O C}$ .

$\Rightarrow C N = R \sin A & O N = R \cos A$ (where $R$ is the circum-radius of $∆ A B C$ ).

Again, the required circle touches the circum-circle internally, whose centre is $K$ .

So, $O K = R - x$ (Difference of their radii). ... $(1)$ .

In triangle $∆ C I D$ , $\frac{I D}{C D} = \tan \frac{C}{2} \Rightarrow C D = \frac{r}{\tan \frac{C}{2}} = \frac{r}{\frac{r}{s - c}} = s - c$

(We have $r = (s - a) \tan \frac{A}{2} = (s - b) \tan \frac{B}{2} = (s - c) \tan \frac{C}{2}$ is the in-radius and $s$ is the semi-perimeter and $a, b & c$ are sides opposite to vertices $A, B & C$ of $∆ A B C$ ).

And $O M = O N + N M = O N + D K = R \cos A + x$ ... $(2)$ .

$K M = N D = C D - C N = (s - c) - \frac{a}{2} = \frac{2 s - 2 c - a}{2} = \frac{a + b + c - 2 c - a}{2} = \frac{b - c}{2}$ ... $(3)$ .

Finally, from right angle triangle $∆ O K M$ , we have $O K^{2} = O M^{2} + K M^{2}$ .

$\Rightarrow {(R - x)}^{2} = {(R \cos A + x)}^{2} + {(\frac{b - c}{2})}^{2}$ (From the above three equations).

$\Rightarrow 2 R x (1 + \cos A) = R^{2} (1 - \cos^{2} A) - {(\frac{b - c}{2})}^{2}$

$\Rightarrow 4 R x \cos^{2} \frac{A}{2} = R^{2} \sin^{2} A - {(\frac{b - c}{2})}^{2}$

$\Rightarrow 4 R x \cos^{2} \frac{A}{2} = {(\frac{a}{2})}^{2} - {(\frac{b - c}{2})}^{2} = (\frac{a + b - c}{2}) (\frac{a - b + c}{2}) = (s - c) (s - b)$

$(∴ \frac{a}{\sin A} = 2 R)$

$\Rightarrow 4 R x \cos^{2} \frac{A}{2} = b c \cdot \frac{(s - c) (s - b)}{b c} = b c \sin^{2} \frac{A}{2}$ $(∵ \sin^{2} \frac{A}{2} = \frac{(s - b) (s - c)}{b c})$ .

$\Rightarrow x = \frac{b c}{4 R} \cdot \frac{\sin^{2} \frac{A}{2}}{\cos^{2} \frac{A}{2}}$ .

$\Rightarrow x = D H = \frac{∆}{a} \tan^{2} \frac{A}{2}$ $(∵ 4 R = \frac{a b c}{∆})$ .

(where, $R$ is circum-radius and $∆$ is the area of $∆ A B C$ ).

If a circle be drawn touching the inscribed and circumscribed circle of a triangle and the side $B C$ , externally, prove that its radius is $\frac{∆}{a} \tan^{2} \frac{A}{2}$ .

Important Questions on Properties of Triangle

Learn and Prepare for any exam you want !

If a circle be drawn touching the inscribed and circumscribed circle of a triangle and the side BC, externally, prove that its radius is ∆atan2A2.

Important Questions on Properties of Triangle

Learn and Prepare for any exam you want !

If a circle be drawn touching the inscribed and circumscribed circle of a triangle and the side $B C$ , externally, prove that its radius is $\frac{∆}{a} \tan^{2} \frac{A}{2}$ .