三角形ABC是等腰直角三角形.自A點做BC垂線交BC于E;則有AE=BE=CE;
可得:AE²+DE²=AD²
BD²=(BE-DE)²=BE²-2BE*DE+DE²
CD²=(CE+DE)²=CE²+2CE*DE+DE²
BD+²CD²=BE²-2BE*DE+DE²+CE²+2CE*DE+DE²
=BE²+DE²+CE²+DE²
=2(AE²+DE²)
=2AD²
看在又打字,又畫圖這么費勁的情況下,給個推薦吧!
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