Question

用勾股定理

Answer 1

证明：(ac+bc)^2=ac^2+bc^2+2ac*bc=ad^2+cd^2+cd^2+bd^2+2ac*bc
ab^2+4cd^2=ad^2+bd^2+2ad*bd+4cd^2
所以只需证明 2ac*bc与 2ad*bd+2cd^2的大小即可
(2ad*bd+2cd^2)/2ac*bc=sinacd*sinbcd+cosacd*cosbcd=cos(∠acd-∠bcd)<0
所以：ad^2+4cd^2>(ac+bc)^2

Answer 2

（AC+BC)²=AC²+BC² +2AC BC
=AC²+CD²+CD²+BC²+2AC BC
= AC²+BC²+2CD²+2AC BC
=AB²+2CD²+2AC BC
（AC+BC)²-AB²-4CD²= AB²+2CD²+2AC BC-AB²+4CD²
=6CD²+2AC BC>0