ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see the given figure). Show that

(i) $\triangle \mathrm{ABE} \cong \triangle \mathrm{ACF}$

(ii) $A B=A C$, i.e., $A B C$ is an isosceles triangle.
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Solution:

(i) In $\triangle \mathrm{ABE}$ and $\triangle \mathrm{ACF}$,

$\angle \mathrm{AEB}=\angle \mathrm{AFC}\left(\right.$ Each $\left.90^{\circ}\right)$

$\angle \mathrm{A}=\angle \mathrm{A}$ (Common angle)

BE $=$ CF (Given)

$\therefore \triangle \mathrm{ABE} \cong \triangle \mathrm{ACF}($ By $\mathrm{AAS}$ congruence rule $)$

(ii) It has already been proved that

$\triangle \mathrm{ABE} \cong \triangle \mathrm{ACF}$

$\Rightarrow \mathrm{AB}=\mathrm{AC}(\mathrm{By} \mathrm{CPCT})$
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