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# ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see figure). Show that(i) $$\triangle{ABE}$$ $$\displaystyle \cong$$ $$\triangle{ACF}$$ (ii) AB = AC i.e., ABC is an isosceles triangle.

In $$\triangle{ABE}$$ and $$\triangle{ACF}$$ , we have,

$$\angle{AEB}$$ = $$\angle{AFC}$$($$\because BE and CF are perpendicular to sides AC and AB) \(\angle{BAE}$$ = $$\angle{CAF}$$ ...($$\because \(\angle{A}$$ is the Common angle)

and BE = CF ...(Given)

$$\therefore$$ $$\triangle{ABE}$$ $$\displaystyle \cong$$ $$\triangle{ACF}$$ ...(By AAS Congruency test)

Thus, AB = AC ...(By CPCT)

$$\therefore$$ $$\triangle{ABC}$$ is an isosceles triangle.
Hence, proved.