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3.  ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see Fig.). Show that these altitudes are equal.


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Consider \Delta AEB   and  \Delta AFC,

(i)     \angle A    is common in both the triangles.

(ii)   \angle AEB\ =\ \angle AFC           (Right angles)

(iii)    AB\ =\ AC              (Given)

Thus by AAS congruence axiom, we can conclude that :

                                 \Delta AEB\ \cong \Delta AFC

Now, by c.p.c.t. we can say :     BE\ =\ CF

Hence these altitudes are equal.

Posted by

Sanket Gandhi

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