Problem
If the midpoint of the side $BC$ of $△ABC$ is $D$, then
$$(1) \ AD<\frac{1}{2} (AB+AC).$$
$$(2) \ AD>\frac{1}{2} (AB+AC-BC).$$
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Solution
$(1)$ If we extend $AD$ and take a point $E$ such that $AD=ED$, we have
$$AD=ED, \quad DC=DB, \quad and \quad ∠ADC=∠EDB,$$
$$∴ \ △ACD≡△EBD,$$
$$∴ \ AC=EB.$$
Regarding $△ABE$,
$$AE<AB+EB,$$
$$∴ \ 2AD<AB+AC,$$
$$∴ \ AD<\frac{1}{2} (AB+AC).$$
$(2)$ Regarding $△ADB$,
$$AD+BD>AB,$$
$$∴ \ AD+\frac{1}{2} BC>AB. \qquad [1]$$
Regarding $△ACD$,
$$AD+CD>AC,$$
$$∴ \ AD+\frac{1}{2} BC>AC. \qquad [2]$$
From $[1]$ and $[2]$,
$$2AD+BC>AB+AC,$$
$$∴ \ 2AD>AB+AC-BC,$$
$$∴ \ AD>\frac{1}{2} (AB+AC-BC).$$
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Reference Teiichiro Sasabe (1976) The Encyclopedia of Geometry (2nd edition), Seikyo-Shinsha, p.23