In the picture, the small (blue) triangle is equilateral. The sides of the large (red) triangle are tangents to the circumcircle of the small triangle at its vertices.

i) Prove that the large triangle is also equilateral and its sides are double that of the small triangle.

ii) Draw this picture, with sides of the smaller triangle 3 centimetres.

iii) Instead of an equilateral triangle, if we draw the tangents to the circumcircle of any other triangle at its vertices, do we get a similar triangle with double the sides? Justify.

Question

Philoid · Accepted Answer

i) Let us label the diagram as shown below,

We know, By alternate segment theorem, angle between chord and tangent is equal to the angle in the other segment.

⇒ ∠RAC = ∠ABC and ∠RCA = ∠ABC [AC is a chord]

As, ΔABC is equilateral

⇒ ∠ABC = 60°

⇒ ∠RAC = ∠RCA = ∠ABC = 60°

Also,

In ΔARC, By angle sum property

∠RAC + ∠RCA + ∠ARC = 180°

⇒ 60° + 60°+ ∠ARC = 180°

⇒ ∠ARC = 60°

⇒ ARC is equilateral triangle and ∠R = 60°

Similarly, we can show

⇒ BPC is equilateral triangle and ∠P = 60°

⇒ AQB is equilateral triangle and ∠Q = 60°

As, ∠P = ∠Q = ∠R = 60°

⇒ ΔPQR is equilateral.

Let, the side of smaller triangle be 'a', i.e. AB = BC = CA = 'a'

But, as ΔARC and ΔAQB are equilateral

⇒ AR = AC = 'a'

And AQ = AB = 'a'

⇒ AQ + AR = a + a

⇒ QR = 2a

As, ΔPQR is equilateral, PQ = QR = PR = '2a'

⇒ Side of larger triangle is double of side of smaller triangle.

ii) Steps of constructions:

1. Draw an equilateral triangle ABC of 3 cm.

2. Draw perpendicular bisectors of sides AB and BC which intersect each other at O.

3. Taking O as center, draw a circle with radius as OA.

4. Draw tangents at Points A, B and C, which makes a triangle PQR.

(iii) The student show try themselves.