Question

A circle touches all three sides of an equilateral triangle ABC at points P,Q,and R . prve that triangle PQR is equilateral triangle​

Answer 1

The triangle PQR is an equilateral triangle, Hence proved.

• Given that, a circle touches all three sides of the equilateral triangle.
• Let, the equilateral triangle be ABC and the circle which touches all three sides be the circle with center at O.
• Therefore, the incircle of the triangle ABC is the same as the centroid of triangle ABC.
• Therefore the points at the intersection of ΔABC and the circle are P,Q and R.
• The segment from the center of circle to P,Q and R will be perpendicular to the segments of the ΔABC.
• The angle formed at the center joining the center and the points P,Q,R is 120° each . As it is an equilateral triangle the points joining the centroid and the perpendicular is always equal to 120°.
• As all the perpendiculars from the circle to the triangle are equal.    (Perpendiculars from the centroid to the triangle are equal in equilateral triangle.)
• The ΔOPR,ΔOQR and ΔOPQ are all isosceles triangles as OP=OQ=OR.
• And angle at O in all the three triangles is 120°.
• Now we find the other two angles in the triangle.
• We consider first ΔOPR,

∠OPR+∠POR+∠ORP = 180°

∠OPR + 120 +∠OPR = 180°.    (Angles opposite to equal sides in the isosceles triangles are equal)

2∠OPR = 60°

∠OPR =30°.

• Similarly for other triangles, the angles will be equal to 30°.Now , the complete angle at P of the ΔPQR will be equal to 30°+30° = 60°.
• Similarly at other points Q and R the angles will be equal to 60°.
• Therefore, the ΔPQR consists of angles 60°,60° and 60°.
• As all the angles of the triangle PQR are equal, the ΔPQR is an equilateral triangle.