Scale factor lab

<p>In the diagram, \(SAT\) is the tangent to the circle at \(A\). \(ADB\) and \(AEC\) are straight lines such that&nbsp;\(B\) and \(C\) are on the circle and \(DE\) is parallel to \(SAT\). Prove that</p>
<div class="graph-block" data-graph-config="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" data-width="217" data-height="151" data-proportional style="width:217px;height:151px;max-width:100%;display:flow-root;margin:0 auto;--graph-width:217px;--graph-aspect:217 / 151;"></div>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">triangles \(ABC\) and \(AED\) are similar, and</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">quadrilateral \(BDEC\) is a cyclic quadrilateral, i.e., there exists a circle that passes through \(B\), \(D\), \(E\) and \(C\).</div></li></ol>

In the diagram, \(SAT\) is the tangent to the circle at \(A\). \(ADB\) and \(AEC\) are straight lines such that \(B\) and \(C\) are on the circle and \(DE\) is parallel to \(SAT\). Prove that

triangles \(ABC\) and \(AED\) are similar, and

quadrilateral \(BDEC\) is a cyclic quadrilateral, i.e., there exists a circle that passes through \(B\), \(D\), \(E\) and \(C\).

Proofs in Plane Geometry - Alternate Segment Theorem

Proofs in Plane Geometry - Alternate Segment Theorem

Video Solution:

Solution:

Solution locked