Parallel-line angles

Scale factor lab

<p>In the diagram, \(TA\) and \(TB\) are tangents from \(T\) to the circle at \(A\) and \(B\), respectively. \(C\) is a point on the circle and \(D\) lies on \(BC\) such that&nbsp;\(CD = AD\). 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">triangles \(TAB\) and \(DAC\) are similar,</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">angle \(ATB + \text{angle}\hspace{0.5em} ADB = 180^\circ\), and</div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">quadrilateral \(ADBT\) is a cyclic quadrilateral, i.e., there exists a circle that passes through \(A\), \(D\), \(B\) and \(T\).</div></li><li data-list-item-id="item-iv" data-list-item-path="item-iv"><div class="list-item-paragraph">Hence, or otherwise, prove that&nbsp;\(TD\) is parallel to \(AC\).</div></li></ol>

In the diagram, \(TA\) and \(TB\) are tangents from \(T\) to the circle at \(A\) and \(B\), respectively. \(C\) is a point on the circle and \(D\) lies on \(BC\) such that \(CD = AD\). Prove that

triangles \(TAB\) and \(DAC\) are similar,

angle \(ATB + \text{angle } ADB = 180^\circ\), and

quadrilateral \(ADBT\) is a cyclic quadrilateral, i.e., there exists a circle that passes through \(A\), \(D\), \(B\) and \(T\).

Proofs in Plane Geometry - Alternate Segment Theorem

Proofs in Plane Geometry - Alternate Segment Theorem

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Solution:

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