maths _35

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.