If a,b,c are the sides of a right angled triangle, where c is hypotenuse, then prove that the radius r of the circle which touches the sides of the triangle is given by $r = \dfrac{{a + b - c}}{2}$.
VerifiedHint:
We can draw a right-angled triangle and a circle with radius r touching the sides of the triangle. Then we can consider the sides of the triangle as the tangents to the circles and equate them using the property that tangents to a circle from an exterior point are equal. Then using the property that radius is perpendicular to the tangent, we can prove the radius with the right angle from a square and equate the sides. Then we can take the hypotenuse and give suitable substitution to make an equation in terms of a, b, c and r. then we can obtain the required relation by rearranging.
Complete step by step solution:
We can draw a right-angled triangle ABC and a circle of radius r centred at O touching the sides of the triangle.
Now we can take the hypotenuse $AC = c$ and let the other sides be $AB = a$ and $BC = b$ .
Consider the lines AE and AG. They are tangents to the circle from the exterior point A. We know that tangents drawn to a circle from an exterior point are equal. Thus, we can write,
$ \Rightarrow AG = AE$ … (1)
Consider the lines BE and BF. They are tangents to the circle from the exterior point B. We know that tangents drawn to a circle from an exterior point are equal. Thus, we can write,
$ \Rightarrow BF = BE$ …. (2)
Consider the lines CF and CG. They are tangents to the circle from the exterior point C. We know that tangents drawn to a circle from an exterior point are equal. Thus, we can write,
$ \Rightarrow CF = CG$ …. (3)
We know that radius is perpendicular to the tangent. From the figure, OE is the radius to the tangent AB and OF is the radius to the tangent BC. So, we can say that $\angle OEB = OFB = 90^\circ $ …. (4)
Also, from the figure, OF and OE is the radius of the same circle. So, they will be equal.
$ \Rightarrow OE = OF = r$ … (5)
Consider the quadrilateral BFOE.
From (2), (4) and (5), we can say that the adjacent sides are equal and 3 angles including $\angle B$ the right angle of the triangles are equal to $90^\circ $ .
So BFOE is a square. For a square, all the sides will be equal.
$ \Rightarrow OE = OF = EB = BF = r$ … (6)
Now we can consider the hypotenuse of the triangle.
$ \Rightarrow c = AC$
From the figure, we can write $AC = AG + GC$
$ \Rightarrow c = AG + GC$
On substituting equations (1) and (3), we get,
$ \Rightarrow c = AE + FC$
From the figure, $FC = BC - BF$ and $AE = AB - BE$ .
$ \Rightarrow c = AB - BE + BC - BF$
On substituting (6), we get,
$ \Rightarrow c = AB - r + BC - r$
Now AB and BC are the non-hypotenuse sides, which are a and b. so we can write,
$ \Rightarrow c = a + b - 2r$
On rearranging, we get,
$ \Rightarrow 2r = a + b - c$
On dividing throughout with 2, we get,
$ \Rightarrow r = \dfrac{{a + b - c}}{2}$
This is the required relation. Hence proved.
Note:
A tangent to a circle is the line that touches exactly one point on a circle. From an exterior point, 2 tangents can be drawn to a circle and their length will be equal. We must draw a diagram and mark all the points. While writing a line as the sum of two lines, we must make sure that we split the line at a point which is located in the line. We cannot simply conclude from the figure that the quadrilateral OEBF is a square. We must prove it by showing that the angles are right angles and adjacent sides are equal.
