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4.  BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.

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Using the given conditions, consider \Delta BEC   and     \Delta CFB,

(i)  \angle BEC\ =\ \angle CFB                 (Right angle)

(ii)    BC\ =\ BC                              (Common in both the triangles)

(iii)   BE\ =\ CF                             (Given that altitudes are of the same length. )

Thus by RHS axiom, we can say that  :                \Delta BEC\ \cong \Delta CFB

Hence by c.p.c.t.,     \angle B\ =\ \angle C

And thus  AB\ =\ AC      (sides opposite to equal angles are also equal). Thus ABC is an isosceles triangle.

                                   

Posted by

Devendra Khairwa

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