F.3 MathsQuadrilaterals!!

(1)
PA=AB=BQ (GIVEN)
AD=AB=BC (PROPERTY OF RHOMBUS)
∴ AP=AD
∠ APD = ∠ ADP (BASE ∠s ,ISOS. △)
AC bisects ∠ CAD (PROPERTY OF RHOMBUS)
∴ ∠ BAC= ∠ CAD
∠BAD = ∠ APD + ∠ ADP (ext. ∠ of △)
∠BAC = ∠ APD
PR // AC (corr. ∠s equal)

similarly

(b)DB//RQ
同上面(a) 一樣,試試自己做

c)
∠DAB + ∠ABC =180ﾟ (PROPERTY OF RHOMBUS)
from answer (a) & (b)
∠DAB = 2∠P
∠ABC = 2∠Q
∴ 2∠P + 2∠Q = 180ﾟ
∠P + ∠Q = 90ﾟ
∠R = 180ﾟ - ∠P - ∠Q = 90ﾟ (∠ sum of △)
angle PRQ is a right angle.

2. AG = AB (PROPERTY of Square).......(1)
CG = CB (PROPERTY of Square)
CD = CF (PROPERTY of Square)
GD = BF ......... .....(2)
∠ G = ∠B = 90ﾟ (PROPERTY of Square).......(3)
△ AGD congrent △ABF (SAS) from 1,2,3
AD=AF (corr. sides congrent △)

AD=DF (GIVEN)
& from answer (a) AD=AF
△ADF is equilateral)
∠ ADF = 60ﾟ (Property of equilateral △)
∠ FDC =45ﾟ (PROPERTY of Square)
∠MDC = 60ﾟ - 45ﾟ =15ﾟ
∠BCG =90ﾟ (PROPERTY of Square)
∠BCD =180ﾟ - 90ﾟ = 90ﾟ (adj. ∠s on st. line)
angle DMF = 180ﾟ - 90ﾟ - 15ﾟ = 75ﾟ (∠ sum of △)