In the given figure,BE and CF …
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Sia ? 5 years, 3 months ago
Given : ABC is a triangle in which altitude BE and CF to side AC and AB are equal.
To Prove : {tex}\triangle \mathrm { ABE } \cong \triangle \mathrm { ACF }{/tex}
Proof : BE = CF ...... [Given]
∠BAE = ∠CAF ...... [Common]
∠AFB = ∠AFC ....... [Each 90o]
{tex}\therefore{/tex} {tex}\triangle \mathrm { ABE } \cong \triangle \mathrm { ACF }{/tex} ........ [By AAS property]
{tex}\therefore{/tex} AB = AC . . .[c.p.c.t.]
{tex}\therefore{/tex} {tex}\triangle{/tex}ABC is an isosceles triangle.
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