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85463295 发表于 2008-1-14 19:34

求角度——等腰直角三角形

等腰Rt△ABC中,∠B为直角,内有一点P,使得AP=1,BP=2,CP=3,求∠APB<br />答案应该是135度,用初中几何知识应该怎么解?<br>

Okppt 发表于 2008-1-14 19:34

将三角形BPC绕点B旋转90度至三角形BP'A处,使BC与BA重合,连结PP',用勾股定理的逆定理证明三角形APP‘是直角三角形<br>

85463295 发表于 2008-1-14 19:34

<img src="http://bbs.pep.com.cn/images/smilies/victory.gif" smilieid="14" border="0" alt="" /><br>

Chelly 发表于 2008-1-14 19:34

谢谢~<br>

Chengqin1987 发表于 2008-1-14 19:34

<br><br>QUOTE:原帖由 南山菊 于 2007-12-29 23:27 发表 <a href="http://219.239.238.42/redirect.php?goto=findpost&pid=3556960&ptid=348491" target="_blank"><img src="http://219.239.238.42/images/common/back.gif" border="0" onload="if(this.width>screen.width*0.7) {this.resized=true; this.width=screen.width*0.7; this.alt='Click here to open new window\nCTRL+Mouse wheel to zoom in/out';}" onmouseover="if(this.width>screen.width*0.7) {this.resized=true; this.width=screen.width*0.7; this.style.cursor='hand'; this.alt='Click here to open new window\nCTRL+Mouse wheel to zoom in/out';}" onclick="if(!this.resized) {return true;} else {window.open(this.src);}" onmousewheel="return imgzoom(this);" alt="" /></a><br />将三角形BPC绕点B旋转90度至三角形BP'A处,使BC与BA重合,连结PP',用勾股定理的逆定理证明三角形APP‘是直角三角形 <br><img src="http://www.oxtoo.com/tmp/0712/54362_676YoUfAC5mdM3Ftr7QR.jpg" border="0" onload="if(this.width>screen.width*0.7) {this.resized=true; this.width=screen.width*0.7; this.alt='Click here to open new window\nCTRL+Mouse wheel to zoom in/out';}" onmouseover="if(this.width>screen.width*0.7) {this.resized=true; this.width=screen.width*0.7; this.style.cursor='hand'; this.alt='Click here to open new window\nCTRL+Mouse wheel to zoom in/out';}" onclick="if(!this.resized) {return true;} else {window.open(this.src);}" onmousewheel="return imgzoom(this);" alt="" /><br>

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