Hướng dẫn giải Bài 98 (Trang 105 SGK Toán 9 Hình học, Tập 2)
<p>Cho đường tròn (O) và một điểm A cố định trên đường tròn. Tìm quỹ tích các trung điểm M của dây AB khi điểm B di động trên đường tròn đó.</p>
<p><strong>Giải :</strong></p>
<p><em>Phần thuận: </em>Giả sử M là trung điểm của dây AB. Ta có <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>O</mi><mi>M</mi><mo>⊥</mo><mi>A</mi><mi>B</mi></math> (định lí).<img class="wscnph" 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width="138" height="146" /></p>
<p>Khi B di động trên (O), điểm M luôn nhìn OA cố định dưới góc vuông, vậy M thuộc đường tròn đường kính OA.</p>
<p><em>Phần đảo: </em>Lấy điểm M' bất kig trên đường tròn đường kính OA.</p>
<p>Nối M' với A, đường thẳng M'A cắt đường tròn (O) tại B'. Nối M' với O ta có <math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mrow><mi>A</mi><mi>M</mi><mo>'</mo><mi>O</mi></mrow><mo>^</mo></mover><mo>=</mo><mn>90</mn><mo>°</mo></math> hay <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>O</mi><mi>M</mi><mo>'</mo><mo>⊥</mo><mi>A</mi><mi>B</mi><mo>'</mo></math> suy ra M' là trung điểm của AB'.</p>
<p><em>Kết luận: </em>Tập hợp các trung điểm của dây AB là đường tròn đường kính OA.</p>
Hướng dẫn Giải Bài 98 (Trang 105, SGK Toán Hình học 9, Tập 2)