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February 10th, 2010, 10:58 PM   #1
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Joined: Nov 2009

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hey guys me again,

Doing circle theorems which is to me, super hard

please show me step by step what you did and how you did it
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 February 11th, 2010, 04:18 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,746 Thanks: 2133 1. Let T be a point "below" P on the given tangent.     Angle SQP = angle TPS by the alternate segment theorem                     = angle PSQ since TP is parallel to SQ,     so SP = PQ (chords subtending equal angles),     and so angle PRS = angle PRQ (angles subtended by equal chords),     i.e., PR bisects ?QRS. 2a. ?CTD + ?CAD = ?CTD + ?TAC + ?TAD                              = ?CTD + ?TCD + ?TDC by the alternate segment theorem                              = 180° (angles of a triangle),       so TCAD is a cyclic quadrilateral (opposite angles supplementary). 2b. ?TAC = ?TCD by the alternate segment theorem                 = ?TAD (both angles subtended by TD, with TCAD cyclic). 2c.            = ?TDC by the alternate segment theorem,       so triangle TCD is isosceles, with TC = TD (equal sides opposite equal angles).
February 11th, 2010, 10:46 AM   #3
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From: Lexington, MA

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Hello, peterle1!

Quote:
 $\text{The diagonals }PR\text{ and }QS\text{ of cyclic quadrilateral }PQRS\text{ intersect at }X.$ $\text{The tangent }TU\text{ at }P\text{ is parallel to {QS.$
Code:
            *           *
*               *
*                 o R
/
*                /  *
*               /   *
S o - - - - - - -o- - o Q
/ X
*           /     *
*         /     *
*      /    *
T - - - - * o * - - - - U
P

Draw chords PQ, QR, RS, and SP.

Quote:
 $\text{(a) Prove that: }\,PQ \,=\,PS$

$\text{Interior angle }QSP\text{ is measured by }\frac{1}{2}\,\text{arc}(PQ)$

$\text{Interior angle }SQP\text{ is measured by }\frac{1}{2}\,\text{arc}(PS)$

$\text{Since }TU\,\parallel\,QS,\;\text{arc}(PQ)\,=\,\text{arc} (PS)$
[color=beige]. . [/color]
Parallel lines intercept equal arcs.

$\text{Hence: }\:\angle QSP \,=\,\angle SQP,\;\text{ and }\Delta PQS\text{ is isosceles.}$

$\text{Therefore: }\;PQ\:=\:PS$

Quote:
 $\text{(b) Prove that: }PR\text{ bisects}\angle QRS$

$\text{Interior angle }PRQ\text{ is measured by }\frac{1}{2}\,\text{arc}(PQ).$

$\text{Interior angle }PRS\text{ is measured by }\frac{1}{2}\,\text{arc}(PS)$

$\text{Since }\text{arc}(PQ) \,=\,\text{arc}(PS),\;\text{then: }\,\angle PRQ \,=\,\angle PRS$

$\text{Therefore, }PR\text{ bisects}\angle QRS.$

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