ABC is a triangle in which BE and CF are, respectively, the perpendiculars to the sides AC and AB. If BE=CF, prove that ΔABC is isosceles.


Given that ABC is a triangle in which BE and CF are perpendiculars to the side AC and AB respectively.


Such that,


BE = CF


We have to prove that, is isosceles triangle.


Now, consider


We have,


BFC = CEB = 90o (Given)


BC = CB (Given)


CF = BE (Given)


So, by RHS congruence rule, we have




FBC = ECB (By c.p.c.t)


ABC = ACB (By c.p.c.t)


AC = AB (Opposite sides of equal angles are equal in a triangle)


Therefore, is isosceles.


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