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. Hint. Show that ,$△BCF$ $\cong △CBE.$

Answer 1

Given $\Delta$ is an isosceles triangle

$\Rightarrow$ AB-BC _______(1)

and $\angle B=\angle C$ ________(2)

Here E andF are midpoints of AC and AB respectively

$\therefore$ AF = FB and AE = EC

know , AB = BC

$\Rightarrow$ AF+FB = AE + EC

$\Rightarrow$ 2AF = 2AE

$\Rightarrow$ AF = AE.

$\Rightarrow$ AF =FB = AE =EC _______(3)

In $\Delta$BCF and $\Delta$ CBE

BC = BC [common side]

$\angle B=\angle C$ [from (2)]

BF = EC [from (3)]

By SAS condition for congruency.

$\Delta BCF\cong \Delta CBE.$

$\therefore$ since $\Delta BCF\cong \Delta$ CBE, by properly of congruncy we can with that

BE = CF.