Bài 7. Trường hợp đồng dạng thứ ba
Hướng dẫn giải Bài 39 (Trang 79 SGK Toán Hình học 8, Tập 2)
<p>Đề b&agrave;i<br />Cho h&igrave;nh thang ABCD (AB//CD). Gọi O l&agrave; giao điểm của hai đường ch&eacute;o AC v&agrave; BD.<br />a) Chứng minh rằng OA.OD=OB.OC.<br />b) Đường thẳng qua O vu&ocirc;ng g&oacute;c với AB v&agrave; CD theo thứ tự tại H v&agrave; K.<br />Chứng minh rằng <math xmlns="http://www.w3.org/1998/Math/MathML"><mstyle displaystyle="false"><mfrac><mrow><mi>O</mi><mo>&#8290;</mo><mi>H</mi></mrow><mrow><mi>O</mi><mo>&#8290;</mo><mi>K</mi></mrow></mfrac><mo>=</mo><mfrac><mrow><mi>A</mi><mo>&#8290;</mo><mi>B</mi></mrow><mrow><mi>C</mi><mo>&#8290;</mo><mi>D</mi></mrow></mfrac></mstyle></math></p> <p><strong class="content_detail">Lời giải chi tiết</strong></p> <p><img src="https://img.loigiaihay.com/picture/2018/0718/b39-trang-79-sgk-toan-8-t2-c2.jpg" /></p> <p>a) V&igrave; $A B //CD (giả thiết)<br />&Aacute;p dụng định l&iacute;:Một đường thẳng cắt hai cạnh của tam gi&aacute;c v&agrave; song song với cạnh c&ograve;n lại tạo th&agrave;nh một tam gi&aacute;c đồng dạng với tam gi&aacute;c đ&atilde; cho.<br /><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>&#8658;</mo><mi mathvariant="normal">&#916;</mi><mo>&#8290;</mo><mi>A</mi><mo>&#8290;</mo><mi>O</mi><mo>&#8290;</mo><mi>B</mi><mo>~</mo><mi mathvariant="normal">&#916;</mi><mo>&#8290;</mo><mi>C</mi><mo>&#8290;</mo><mi>O</mi><mo>&#8290;</mo><mpadded><mi>D</mi></mpadded><mo>&#8290;</mo></mrow><mspace linebreak="newline"/><mo>&#8658;</mo><mfrac><mstyle displaystyle="true"><mi>O</mi><mo>&#8290;</mo><mi>A</mi></mstyle><mstyle displaystyle="true"><mi>O</mi><mo>&#8290;</mo><mi>C</mi></mstyle></mfrac><mo>=</mo><mfrac><mstyle displaystyle="true"><mi>O</mi><mo>&#8290;</mo><mi>B</mi></mstyle><mstyle displaystyle="true"><mi>O</mi><mo>&#8290;</mo><mi>D</mi></mstyle></mfrac><mo>&#160;</mo><mo>(</mo><mi>c</mi><mi>&#7863;</mi><mi>p</mi><mo>&#160;</mo><mi>c</mi><mi>&#7841;</mi><mi>n</mi><mi>h</mi><mo>&#160;</mo><mi>t</mi><mi>&#432;</mi><mi>&#417;</mi><mi>n</mi><mi>g</mi><mo>&#160;</mo><mi>&#7913;</mi><mi>n</mi><mi>g</mi><mo>&#160;</mo><mi>t</mi><mi>&#7881;</mi><mo>&#160;</mo><mi>l</mi><mi>&#7879;</mi><mo>)</mo><mspace linebreak="newline"/><mo>&#8658;</mo><mi>O</mi><mi>A</mi><mo>.</mo><mi>O</mi><mi>D</mi><mo>=</mo><mi>O</mi><mi>C</mi><mo>.</mo><mi>O</mi><mi>B</mi></math><br />b) Theo c&acirc;u a) ta c&oacute; <math xmlns="http://www.w3.org/1998/Math/MathML"><mstyle displaystyle="false"><mi mathvariant="normal">&#916;</mi><mo>&#8290;</mo><mi>A</mi><mo>&#8290;</mo><mi>O</mi><mo>&#8290;</mo><mi>B</mi><mo>&#8290;</mo><mo>~</mo><mo>&#8290;</mo><mi mathvariant="normal">&#916;</mi><mo>&#8290;</mo><mi>C</mi><mo>&#8290;</mo><mi>O</mi><mo>&#8290;</mo><mi>D</mi></mstyle></math> n&ecirc;n <math xmlns="http://www.w3.org/1998/Math/MathML"><mstyle displaystyle="false"><mfrac><mrow><mi>O</mi><mo>&#8290;</mo><mi>A</mi></mrow><mrow><mi>O</mi><mo>&#8290;</mo><mi>C</mi></mrow></mfrac><mo>=</mo><mfrac><mrow><mi>A</mi><mo>&#8290;</mo><mi>B</mi></mrow><mrow><mi>C</mi><mo>&#8290;</mo><mi>D</mi></mrow></mfrac></mstyle></math><br />X&eacute;t <math xmlns="http://www.w3.org/1998/Math/MathML"><mstyle displaystyle="false"><mi mathvariant="normal">&#9651;</mi><mo>&#8290;</mo><mi>A</mi><mo>&#8290;</mo><mi>O</mi><mo>&#8290;</mo><mi>H</mi></mstyle></math> v&agrave; <math xmlns="http://www.w3.org/1998/Math/MathML"><mstyle displaystyle="false"><mi mathvariant="normal">&#9651;</mi><mo>&#8290;</mo><mi>C</mi><mo>&#8290;</mo><mi>O</mi><mo>&#8290;</mo><mi>K</mi></mstyle></math> c&oacute;:<br /><math xmlns="http://www.w3.org/1998/Math/MathML"><mstyle displaystyle="false"><mover accent="true"><mrow><mi>A</mi><mo>&#8290;</mo><mi>H</mi><mo>&#8290;</mo><mi>O</mi></mrow><mo>^</mo></mover><mo>=</mo><mover accent="true"><mrow><mi>C</mi><mo>&#8290;</mo><mi>K</mi><mo>&#8290;</mo><mi>O</mi></mrow><mo>^</mo></mover><mo>=</mo><msup><mn>90</mn><mo>&#8728;</mo></msup></mstyle></math><br /><math xmlns="http://www.w3.org/1998/Math/MathML"><mstyle displaystyle="false"><mover accent="true"><mrow><mi>H</mi><mo>&#8290;</mo><mi>O</mi><mo>&#8290;</mo><mi>A</mi></mrow><mo>^</mo></mover><mo>=</mo><mover accent="true"><mrow><mi>K</mi><mo>&#8290;</mo><mi>OC</mi></mrow><mo>^</mo></mover></mstyle></math> (đối đỉnh)<br /><math xmlns="http://www.w3.org/1998/Math/MathML"><mstyle displaystyle="false"><mo>&#8658;</mo><mi mathvariant="normal">&#916;</mi><mo>&#8290;</mo><mi>A</mi><mo>&#8290;</mo><mi>O</mi><mo>&#8290;</mo><mi>H</mi><mo>&#8290;</mo><mo>~</mo><mi mathvariant="normal">&#916;</mi><mo>&#8290;</mo><mi>C</mi><mo>&#8290;</mo><mi>O</mi><mo>&#8290;</mo><mi>K</mi><mo>&#8290;</mo><mrow><mo>(</mo><mi>g</mi><mo>-</mo><mi>g</mi><mo>)</mo></mrow></mstyle></math><br /><math xmlns="http://www.w3.org/1998/Math/MathML"><mstyle displaystyle="false"><mo>&#8658;</mo><mfrac><mrow><mi>O</mi><mo>&#8290;</mo><mi>H</mi></mrow><mrow><mi>O</mi><mo>&#8290;</mo><mi>K</mi></mrow></mfrac><mo>=</mo><mfrac><mrow><mi>O</mi><mo>&#8290;</mo><mi>A</mi></mrow><mrow><mi>O</mi><mo>&#8290;</mo><mi>C</mi></mrow></mfrac></mstyle></math> (2) (cặp cạnh tương ứng tỉ lệ)<br />Từ (1) v&agrave; (2) <math xmlns="http://www.w3.org/1998/Math/MathML"><mstyle displaystyle="false"><mo>&#8658;</mo><mfrac><mrow><mi>O</mi><mo>&#8290;</mo><mi>H</mi></mrow><mrow><mi>O</mi><mo>&#8290;</mo><mi>K</mi></mrow></mfrac><mo>=</mo><mfrac><mrow><mi>A</mi><mo>&#8290;</mo><mi>B</mi></mrow><mrow><mi>C</mi><mo>&#8290;</mo><mi>D</mi></mrow></mfrac></mstyle></math></p>
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