circle geometry (1 Viewer)

coyazayo · Apr 12, 2016

Hi everyone reading this thread and thanks in advance

InteGrand · Apr 12, 2016

coyazayo said:
Hi everyone reading this thread and thanks in advance

$\noindent These proofs will be based on the fact that equal arcs in a circle subtend equal angles at the circumference and centre (and the converse of this).$

$\noindent i) Note that $\angle PAX = \angle QAY$ (vertically opposite angles). So these are two equal angles subtended at the circumference of congruent circles, which implies that arcs subtending them are equal, so arc $PX$ equals arc $QY$.$

$\noindent ii) Note that the two circles' minor arcs $AB$ are equal. This can be proved by letting the left circle have centre $O$, and the right have centre $O^\prime$, and then showing that $AOB \equiv \triangle AO^\prime B$ using SSS (as the circles have the same radius). It follows that $\angle AOB = \angle AO^\prime B$ by corresponding angles in congruent triangles. This implies that the two arcs are equal, as they subtend equal angles in congruent circles' centres.$

$\noindent So since these arcs are equal, $\angle BPA = \angle BQA = \alpha$ say (equal arcs in congruent circles subtend equal angles at the circumference). But as $PAQ$ is a straight line, $\angle BPQ = \angle BPA = \alpha$ and similarly $\angle BPQ = \angle BPA = \alpha$. So $BP=BQ$ (sides opposite equal angles in $\triangle BPQ$).$

$\noindent (iii) Note that $\angle PBX = \angle YBQ$, since the arcs $PX$ and $QY$ are equal in congruent circles, so they subtend equal angles at the circumference. It follows that$

$\begin{align*}\angle PBQ &= \angle PBX + \angle XBQ \quad (\text{adjacent angles}) \\ &= \angle YBQ + \angle XBQ \\ &= \angle XBY \quad (\text{adjacent angles}).\end{align*}$

circle geometry (1 Viewer)

coyazayo

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InteGrand

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