Scale factor lab

<p>In the diagram, \(TA\) is the tangent to the circle at \(A\) and \(BC\) is a diameter of the circle. \(S\) is a point on \(TA\) produced and \(OS\) intersects \(AB\) at \(M\). The circle that passes through \(A\), \(M\) and \(S\) has its centre at the midpoint of \(AS\). Prove that</p>
<div class="graph-block" data-graph-config="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" data-width="288" data-height="193" data-proportional style="width:288px;height:193px;max-width:100%;display:flow-root;margin:0 auto;--graph-width:288px;--graph-aspect:288 / 193;"></div>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">triangles \(ABT\) and \(CAT\) are similar,</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">triangles \(AMS\) and \(CAB\) are similar, and</div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">quadrilateral \(ASOC\) is a trapezium.</div></li></ol>

In the diagram, \(TA\) is the tangent to the circle at \(A\) and \(BC\) is a diameter of the circle. \(S\) is a point on \(TA\) produced and \(OS\) intersects \(AB\) at \(M\). The circle that passes through \(A\), \(M\) and \(S\) has its centre at the midpoint of \(AS\). Prove that

Proofs in Plane Geometry - Alternate Segment Theorem

Proofs in Plane Geometry - Alternate Segment Theorem

Video Solution:

Solution:

Solution locked