Question
In the adjoining figure, $\triangle\text{ABC}$ is an isosceles triangle in which $AB = AC.$ If $E$ and $F$ be the midpoints of $AC$ and $AB$ respectively, prove that $BE = CF.$
✓
Answer
Given: $\triangle\text{ABC}$ is an isosceles triangle in which $AB = AC.\ E$ and $F$ are the midpoints of $AC$ and $AB$ respectively.
To prove: $BE = CF$ Proof: IN $\triangle\text{BCF}$ and $\triangle\text{CBE},$
$BC = BC ($common$) BF = CE$
$($Half of equal sides $AB$ and $AC)$
$\angle\text{CBF}=\angle\text{BCF}$ (Angles opposite to equal sides) $\triangle\text{BCF}\cong\triangle\text{CBE}$
$(SAS$ condition$)$
$CF = BE (c.p.c.t.)$ or $BE = CF$ Hence proved.
To prove: $BE = CF$ Proof: IN $\triangle\text{BCF}$ and $\triangle\text{CBE},$
$BC = BC ($common$) BF = CE$
$($Half of equal sides $AB$ and $AC)$
$\angle\text{CBF}=\angle\text{BCF}$ (Angles opposite to equal sides) $\triangle\text{BCF}\cong\triangle\text{CBE}$
$(SAS$ condition$)$
$CF = BE (c.p.c.t.)$ or $BE = CF$ Hence proved.
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