Asked by masara | 28th Dec, 2009, 02:56: PM
Let AD, BE and CF be the altitudes.
Then in BFC and BEC
BC is common
BFC = BEC (Each 90o)
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.
Answered by | 28th Dec, 2009, 03:30: PM
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