Scale factor lab

<p>In the diagram, \(ABCD\) is a cyclic quadrilateral. \(AB\) is a diameter and is parallel to \(CD\). \(TDS\) and \(TBR\) are tangents to the circle at \(D\) and \(B\), respectively. Given that&nbsp;\(TDS\) intersects \(OA\) produced at \(S\), prove that</p>
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<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">angle \(ADS =\) angle \(TBC\) ,</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">angle \(CBD = 90^\circ - 2 \times\) angle \(SDA\) , and</div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">triangles \(SAD\) and \(BCD\) are similar.</div></li></ol>

In the diagram, \(ABCD\) is a cyclic quadrilateral. \(AB\) is a diameter and is parallel to \(CD\). \(TDS\) and \(TBR\) are tangents to the circle at \(D\) and \(B\), respectively. Given that \(TDS\) intersects \(OA\) produced at \(S\), prove that

Proofs in Plane Geometry - Alternate Segment Theorem

Proofs in Plane Geometry - Alternate Segment Theorem

Video Solution:

Solution:

Solution locked