Question

The circle in the figure below touches all the three lines.

Prove that the perimeter of the right angle triangle is equal to the diameter of the circle.

Answer 1

Construction: Joining the centre O of the circle with the point at which the tangents touch the circle.

After naming, we have the following figure.

We know that any tangent to circle is perpendicular to the radius to the point of contact.

AE is the tangent to the circle and OE is the radius of the circle.

∴ ∠OEA = 90°

Similarly, ∠AFO = 90°

In quadrilateral AFOE:

∠A = ∠OEA = ∠OFA = 90°

Thus, by angle sum property of quadrilaterals:

∠EOF = 90°

Also, OE = OF = r (say) (Radii of the same circle)

As all angles of quadrilateral OEAF are of measure 90° and adjacent sides are equal, quadrilateral OEAF is a square.

We know that lengths of tangents drawn from a point outside a circle are equal.

⇒ AE = AF

⇒ AE = AF = OE = OF = r (All sides of a square are equal in length)

Now, AE + AF = r + r

= 2r

= d (where d is the diametre of the circle) …. (1)

Also, DE = DC and BC = BF (Lengths of tangents drawn from a point outside a circle) …(2)

We know that perimeter of a triangle is equal to the sum of the lengths of its sides.

∴ Perimeter of ΔDAB = AD + DB + BA

= AD + DC + CB + BA

= AD + DE + BF + BA (From equation (2))

= AE + AF

= d (From equation (1))

Therefore, the perimeter of the right-angled triangle is equal to the diametre of the circle.