$已知：?△ABC?中，?BC=a?，?AC=b?，?AB=c?，?a^2+b^2=c^2?$
$求證：?△ABC?為直角三角形$
$證明如下：作一直角三角形? D E F?，?∠F=90°?，使其兩直角邊與三角形?ABC?$
$的兩條較短邊相等，即?EF=BC=a?，?DF=AC=b?$
$∵?Rt△DEF?，?∠F=90°?$
$∴?DE2=EF2+DF2=a2+b2=c2=AB2?$
$∴?DE=AB?$
$在?△DEF ?與?△ABC?中$
$?\begin {cases}{D E=A B}\\{D F=A C}\\{ E F=B C}\end {cases}?$
$∴?△DEF≌△ABC(\mathrm {SSS})?$
$∴?∠C=∠F=90°?$
$∴?△ABC?為直角三角形$
