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maths _35
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maths _35 [IMG]http://i293.photobucket.com/albums/mm67/zaza520/0004-58.jpg[/IMG]
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Angle CBA = Angle BAE (int. angle of reg. pentagon). CBAE is a cyclic quad. so angle CBA = angle PEA ( ext. angle of cyclic quad.) therefore, angle BAE = angle PEA. For triangle EPM and triangle BMA EM = MA ( M is the mid-point of EA, given) angle PEA = angle BAE ( proved) angle EMP = angle BMA ( vert. opposite angles) therefore triangle EPM congruent triangle BMA (ASA) therefore PM = MB. For triangle EMB and triangle PAM EM = MA ( given) angle EMB = angle PMA ( vert. opposite angles) MB = PM ( proved) therefore, triangle EMB congruent triangle PAM ( SAS). also, angle BEM = angle MAP. (2) Angle EBA = angle BEA ( base angles of isos. triangle EAB, EA = BA) therefore angle MAP = angle BEM = angle EBA therefore PA is tangent to circle ( angle in alt. segment equal) (3) EPAB is a parallelogram ( based on the results of the above 2 pairs of congruent triangles). therefore EB//PA. Arc CB = arc DE ( ABCDE is a reg. pentagon) therefore angle CDB = angle DBE ( angles subtends equal arc) therefore CD//EB ( alt. angles equal) therefore PA//EB//CD.
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