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Question

# In the adjoining figure, ∆ABC is an isosceles triangle in which AB = AC. If E and F be the midpoints of AC and AB respectively, prove that BE = CF. Figure

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Solution

## $Given:\phantom{\rule{0ex}{0ex}}△ABC\mathrm{is}\mathrm{an}\mathrm{isosceles}\mathrm{triangle}\mathrm{in}\mathrm{which}AB=AC.\phantom{\rule{0ex}{0ex}}E\mathrm{and}F\mathrm{are}\mathrm{midpoints}\mathrm{of}AC\mathrm{and}AB,\mathrm{respectively}.\phantom{\rule{0ex}{0ex}}\mathrm{To}\mathrm{prove}:\phantom{\rule{0ex}{0ex}}BE=CF\phantom{\rule{0ex}{0ex}}\mathrm{Proof}:\phantom{\rule{0ex}{0ex}}E\mathrm{and}F\mathrm{are}\mathrm{midpoints}\mathrm{of}AC\mathrm{and}AB,\mathrm{respectively}.\phantom{\rule{0ex}{0ex}}=>AF=FB,AE=EC\phantom{\rule{0ex}{0ex}}AB=AC\phantom{\rule{0ex}{0ex}}=>\frac{1}{2}AB=\frac{1}{2}AC\phantom{\rule{0ex}{0ex}}=>FB=EC\phantom{\rule{0ex}{0ex}}\angle ABC=\angle ACB\left(\mathrm{angle}\mathrm{opposite}\mathrm{to}\mathrm{equal}\mathrm{sides}\mathrm{are}\mathrm{equal}\right)\phantom{\rule{0ex}{0ex}}=>\angle FBC=\angle ECB\phantom{\rule{0ex}{0ex}}\mathrm{Consider}△BCF\mathrm{and}△CBE:\phantom{\rule{0ex}{0ex}}BC=BC\left(\mathrm{common}\right)\phantom{\rule{0ex}{0ex}}\angle FBC=\angle ECB\left(\mathrm{proved}\mathrm{above}\right)\phantom{\rule{0ex}{0ex}}FB=EC\left(\mathrm{proved}\mathrm{above}\right)\phantom{\rule{0ex}{0ex}}BySAS\mathrm{congruence}\mathrm{property}:\phantom{\rule{0ex}{0ex}}△BCF\cong △CBE\phantom{\rule{0ex}{0ex}}BE=CF\left(\mathrm{corresponding}\mathrm{parts}\mathrm{of}\mathrm{the}\mathrm{congruent}\mathrm{triangles}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

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