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Question
BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.
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Solution
In ΔBEC and ΔCFB,
∠BEC = ∠CFB ...(Each 90°)
BC = CB ...(Common)
BE = CF ...(Given)
∴ ΔBEC ≅ ΔCFB ...(By RHS congruence rule)
⇒ ∠BCE = ∠CBF ...(By Corresponding parts of congruent triangles)
Now, in ΔABC, ∠BCA = ∠CBA
∴ AB = AC ...(Sides opposite to equal angles of a triangle are equal)
Hence, ΔABC is isosceles.
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