Question

In the given triangle $ABC$, $D,E$ and $F$ are the mid-points of sides $BC,CA$ and $AB$ respectively. Prove that

$\frac{AB-BC}{2}<AE<\frac{AB+BC}{2}.$

Answer 1

We know that difference of two sides is less than the third side

$\therefore AB-BC<AC$

Since $E$ is the mid-point of $AC$, $AE=\frac{AC}{2}$

$AB-BC<2AE$

$\frac{AB-BC}{2}<AE.....\left(1\right)$

We know that the sum of the two sides is greater than the third side

$AB+BC>AC.AC=2AE.AB+BC>2AE$

$AE<\frac{AB+BC}{2}.....\left(2\right)$

From (1) ad (2),

$\frac{AB-BC}{2}<AE<\frac{AB+BC}{2}.$ $\left[henceproved\right]$