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प्रश्न
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.
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उत्तर
△ABC is an isosceles triangle.
∴ AB = AC
∠ACB = ∠ABC ...[Angles opposite to equal sides of a △ are equal]
∠BCE = ∠CBF
Now, in △BEC and △CFB
∠BCE = ∠CBF ...[Proved above]
∠BEC = ∠CFB ...[Each 90°]
BC = CB ...[Common]
∴ △BEC ≅ △CFB ...[By AAS congruence]
So, BE = CF ...[By Corresponding parts of congruent triangles]
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संबंधित प्रश्न
ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see the given figure). Show that
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- AB = AC, i.e., ABC is an isosceles triangle.
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In ∆ABD and ∆ACD,
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