Question
$B E$ and $C F$ are two equal altitudes of a triangle $A B C$. Using $RHS$ congruence rule, prove that a triangle $A B C$ is isosceles.
✓
Answer
Given: $BE $and $CF $are two equal altitudes of a triangle $ABC .$
To Prove: $\triangle A B C$ is isosceles.
Proof : In right $\triangle B E C$ and right $\triangle C F B$
side $B E=$ side $C F$...[Given]
$B C=C B$...[Common]
$\triangle B E C=\triangle C F B \ldots[B y \text { RHS rule }]$
$\therefore \angle BCE=\angle CBF \ldots[\text { c.p.c.t.] }$
$\therefore A B=A C$...[Sides opposite to equal angles of a triangle are equal]
$\therefore \triangle ABC$ is isosceles.
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