ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see the given figure). Show that these altitudes are equal.
![](/media/uploads/2021/12/ABC-is-an-isosceles-triangle-in-which-altitudes-BE-and-CF-are-drawn-to-equal-sides.png)
Solution:
In $\triangle \mathrm{AEB}$ and $\triangle \mathrm{AFC}$,
$\angle A E B$ and $\angle A F C\left(\right.$ Each $\left.90^{\circ}\right)$
$\angle \mathrm{A}=\angle \mathrm{A}$ (Common angle)
$\mathrm{AB}=\mathrm{AC}$ (Given) \ $\therefore \triangle \mathrm{AEB} \cong \triangle \mathrm{AFC}($ By $\mathrm{AAS}$ congruence rule $)$
$\Rightarrow \mathrm{BE}=\mathrm{CF}(\mathrm{By} \mathrm{CPCT})$
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