<p>In the diagram, \(TA\) and \(TC\) are tangents from \(T\) to the circle at \(A\) and \(C\), respectively. \(B\) is a point on the circle and the tangent to the circle at \(B\) intersects \(TC\) and \(TA\) at \(D\) and \(E\), respectively. 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 \(EDT = 2 \times \text{angle}\hspace{0.5em} BAC\) ,</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">angle \(ABC = 90^\circ + \frac{1}{2} \times \text{angle}\hspace{0.5em} DTE\) , and</div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">quadrilateral \(ACDB\) is NOT a cyclic quadrilateral, i.e., there exists no such a circle that passes through \(A\), \(C\), \(D\) and \(B\).</div></li></ol>

In the diagram, \(TA\) and \(TC\) are tangents from \(T\) to the circle at \(A\) and \(C\), respectively. \(B\) is a point on the circle and the tangent to the circle at \(B\) intersects \(TC\) and \(TA\) at \(D\) and \(E\), respectively. Prove that

angle \(EDT = 2 \times \text{angle } BAC\) ,

angle \(ABC = 90^\circ + \frac{1}{2} \times \text{angle } DTE\) , and

quadrilateral \(ACDB\) is NOT a cyclic quadrilateral, i.e., there exists no such a circle that passes through \(A\), \(C\), \(D\) and \(B\).

Proofs in Plane Geometry - Alternate Segment Theorem

Proofs in Plane Geometry - Alternate Segment Theorem

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