In figure, $D E F$ is an isosceles triangle and $D G$ is the altitude on $F E$ is $D G$ also a median.

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Solution

$D F = D E$ (Isoceles)
$D G = D G$ (common side)
$∠ D G F = ∠ D G E (90^{0})$
Hence $Δ D G F$ and $Δ D G E$ are congruent.
$F G = G E$
Hence $D G$ is a median.

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In figure, DEF is an isosceles triangle and DG is the altitude on FE is DG also a median.

DF=DE (Isoceles)DG=DG (common side)∠DGF=∠DGE(900)Hence ΔDGF and ΔDGE are congruent.FG=GEHence DG is a median.

In figure, $D E F$ is an isosceles triangle and $D G$ is the altitude on $F E$ is $D G$ also a median.

$D F = D E$ (Isoceles)
$D G = D G$ (common side)
$∠ D G F = ∠ D G E (90^{0})$
Hence $Δ D G F$ and $Δ D G E$ are congruent.
$F G = G E$
Hence $D G$ is a median.