<p>In the diagram, \(TAR\) is the tangent to the circle at \(A\). \(DCA\) and \(DOB\) are straight lines. It is given that angle \(BAR = 3 \times \text{angle}\hspace{0.5em} OCB\), prove that</p>
<div class="graph-block" data-graph-config="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" data-width="245" data-height="168" data-proportional style="width:245px;height:168px;max-width:100%;display:flow-root;margin:0 auto;--graph-width:245px;--graph-aspect:245 / 168;"></div>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">angle \(ODC = 2 \times \text{angle}\hspace{0.5em} OBC\),</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">triangle \(COD\) is an isosceles triangle, and</div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">angle \(DCO = \text{angle}\hspace{0.5em} CAT + 90^\circ\).</div></li></ol>

In the diagram, \(TAR\) is the tangent to the circle at \(A\). \(DCA\) and \(DOB\) are straight lines. It is given that angle \(BAR = 3 \times \text{angle } OCB\), prove that

angle \(ODC = 2 \times \text{angle } OBC\),

triangle \(COD\) is an isosceles triangle, and

angle \(DCO = \text{angle } CAT + 90^\circ\).

Proofs in Plane Geometry - Alternate Segment Theorem

Proofs in Plane Geometry - Alternate Segment Theorem

Video Solution:

Solution:

Solution locked