CBSE Class 9 Answered

Let AD, BE and CF be the altitudes.

Then in BFC and BEC

BC is common

BFC = BEC         (Each 90^o)

FC = BE (altitudes are equal)
BFC BEC                (RHS)

FBC = ECB            (By c.p.c.t)

Therefore, ABC = ACB          

i.e AB = AC        (Sides opposite to equal angles)

Similarly, ADC AFC

FAC = DCA            (By c.p.c.t)

Therefore, BAB = BCA          

i.e AB = BC      (Sides opposite to equal angles)

Hence, AB = BC = AC.

So, the triangle is an equilateral triangle.