解：當(dāng)?$\triangle ABC$?是鈍角三角形時(shí)，猜想?$a^2+b^2＜c^2$?
證明：過點(diǎn)?$A$?作?$AD\perp BC$?的延長線于點(diǎn)?$D$?
設(shè)?$CD = x$?
?$ $?在?$Rt\triangle ABD$?中，?$AD^2=c^2-(a+x)^2$?
?$ $?在?$Rt\triangle ADC$?中，?$AD^2=b^2-x^2$?
∴?$c^2-(a+x)^2=b^2- x^2$?
化簡得?$a^2+b^2=c^2-2ax$?
∵?$a＞0，$??$x＞0，$?∴?$2ax＞0$?
∴?$a^2+b^2＜c^2$?
