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

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

$$\angle{AEB}$$ = $$\angle{AFC}$$ ...(BE and CF are perpendiculars drawn to sides AC and AB respectively)

Also, $$\angle{A}$$ is common angle.
and AB = AC ...(Given)

Therefore, $$\triangle{ABE}$$ $$\displaystyle \cong$$ $$\triangle{ACF}$$ ...(By AAS congruency test)

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

Hence, BE = CF ...(By CPCT)
Hence, proved.