3 maths questions

1, (a)
AB//FH (given)
BH = HD (properties of //gram)
AF = FD (Mid-pt Theorem)
angle HAD = angle ADE (alt angles, // lines)
angle AFG = angle EFD (vertical opp. angles)
triangle AFG is congruent to triangle EFD (ASA)
Therefore, ED = GA (corresponding line, congruent triangles)
(b)
ED = GA
ED = 2GH
AH = GA + GH = 1.5 ED
Therefore, ED : AH = 2 : 3.

2.
AP : AB = AQ : AC (corr. sides, similar triangles)
AR : AD = AQ : AC (corr. sides, similar triangles)
Therefore, AP : AB = AR : AD
Furthermore, angle PAR = angle BAD (common angle)
Therefore triangle APR ~ triangle ABD (2 sides proportion
Therefore, PR//BD (corresponding angle equal)

3.(a)
ADMN and BCMN are rectangle.
Therefore, AD//MN//BC.
(b)
MN//BC
PR = RS ( Mid point theorem)
angle PRB = 90 = angle SRB
BR = BR (common side)
Therefore, triangle PRB is congruent with triangle SRB. (RHS)
(c)
BP = BP (common side)
AP = PH (given)
angle BAP = angle BRP (proved)
Similarly using RHS, triangle ABP is congruent with triangle RBP.
angle PBS = 2*90/3 = 60
angle RSB = 180 - 30 - 90 = 60
angle RPS = 60
Therefore triangle PBS is an equilateral triangle