解:因?yàn)?img class="mathml" src="http://math.21cnjy.com/MathMLToImage?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmi%3ED%3C%2Fmi%3E%3Cmi%3EE%3C%2Fmi%3E%3Cmn%3E%2F%3C%2Fmn%3E%3Cmn%3E%2F%3C%2Fmn%3E%3Cmi%3EA%3C%2Fmi%3E%3Cmi%3EC%3C%2Fmi%3E%3C%2Fmath%3E" style="max-width:100%;vertical-align: middle;">(已知),
所以 ▲ ( ).
因?yàn)?img class="mathml" src="http://math.21cnjy.com/MathMLToImage?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmo%3E%E2%88%A0%3C%2Fmo%3E%3Cmi%3EB%3C%2Fmi%3E%3Cmi%3EE%3C%2Fmi%3E%3Cmi%3ED%3C%2Fmi%3E%3Cmo%3E%3D%3C%2Fmo%3E%3Cmo%3E%E2%88%A0%3C%2Fmo%3E%3Cmi%3ED%3C%2Fmi%3E%3Cmi%3EF%3C%2Fmi%3E%3Cmi%3EC%3C%2Fmi%3E%3C%2Fmath%3E" style="max-width:100%;vertical-align: middle;">(已知),
所以 ▲ (等量代換)
所以( ).
所以( ).
解:因?yàn)?img class="mathml" src="http://math.21cnjy.com/MathMLToImage?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmtext%3E%E2%88%A0%3C%2Fmtext%3E%3Cmi%3EA%3C%2Fmi%3E%3Cmi%3EB%3C%2Fmi%3E%3Cmi%3EC%3C%2Fmi%3E%3Cmo%3E%3D%3C%2Fmo%3E%3Cmo%3E%E2%88%A0%3C%2Fmo%3E%3Cmi%3EA%3C%2Fmi%3E%3Cmi%3EC%3C%2Fmi%3E%3Cmi%3EB%3C%2Fmi%3E%3C%2Fmath%3E" style="max-width:100%;vertical-align: middle;">(已知),
所以( ).
又因?yàn)?img class="mathml" src="http://math.21cnjy.com/MathMLToImage?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmo%3E%E2%88%A0%3C%2Fmo%3E%3Cmn%3E1%3C%2Fmn%3E%3Cmo%3E%3D%3C%2Fmo%3E%3Cmo%3E%E2%88%A0%3C%2Fmo%3E%3Cmn%3E2%3C%2Fmn%3E%3C%2Fmath%3E" style="max-width:100%;vertical-align: middle;">(已知).
所以 ▲ ▲ (等式性質(zhì))
所以
在和中,
,
所以( )(完成以下說(shuō)理過(guò)程)
①在直線a上取一點(diǎn)A,作直線(與直線a不垂直);
②在的延長(zhǎng)線上取一點(diǎn)B,以B為圓心長(zhǎng)為半徑作弧,交直線a于點(diǎn)C;
③聯(lián)結(jié) , 以B為圓心長(zhǎng)為半徑作弧,交于點(diǎn)D,作直線
這樣,就得到直線 . 你能說(shuō)明的理由嗎?
①試說(shuō)明∠OAP=∠QPB的理由,
②如果△BPQ是等腰三角形,求出點(diǎn)Q的坐標(biāo).