Scale factor lab

<p>In the diagram, \(BC\) is a diameter of the circle. Triangle \(BCR\) is an isosceles triangle with \(BC = BR\) and \(CR\) intersects the circle at \(A\). The tangent to the circle at \(A\) intersects \(BR\) 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 \(BSA = 90^\circ\) ,</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">triangles \(ABC\) and \(SBA\) are similar, and</div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">the circle that passes through \(A\), \(B\) and \(S\) has its centre at the midpoint of \(AB\).</div></li></ol>

In the diagram, \(BC\) is a diameter of the circle. Triangle \(BCR\) is an isosceles triangle with \(BC = BR\) and \(CR\) intersects the circle at \(A\). The tangent to the circle at \(A\) intersects \(BR\) at \(S\). Prove that

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

the circle that passes through \(A\), \(B\) and \(S\) has its centre at the midpoint of \(AB\).

Proofs in Plane Geometry - Alternate Segment Theorem

Proofs in Plane Geometry - Alternate Segment Theorem

Video Solution:

Solution:

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