Asked by | 9th Jan, 2009, 07:30: PM
For an equilateral triangle ABC,
Consider that D is the mid point of BC.
Join A and D.
Now, In triangles ABD and ACD,
Angle B = Angle C(each is equal to 60 degrees)
BD=DC(D is the mid point of BC)
AB=AC((Sides of an equilateral triangle)
Thus, triangle ABD is congruent to triangle ACD(SAS rule)
So angle ADB=angle ACD(C.P.C.T.)
But, they form a linear pair.So each of them must be equal to 90 degrees.
angle BAD =angle CAD(C.P.C.T.)
Thus, we see that the line segment AD which was a median originally has turned out be an angle bisector and also a parprndicular bisector of the triangle ABC.
So, we can conclude that a median of an equilateral triangle is an angle bisector as well as a perpendicular bisector.
Since centroid is the point of intersection of the three medians for any triangle, in this case it also becomes the point of intersection of the three perpendicular bisectors.Which means that it's also the circumcenter of the given equilateral triangle.(as the point of intersection of the perpendicular bisectors of a triangle is the circumcenter)
Answered by | 13th Jan, 2009, 05:41: PM
Kindly Sign up for a personalised experience
- Ask Study Doubts
- Sample Papers
- Past Year Papers
- Textbook Solutions
Verify mobile number
Enter the OTP sent to your number