Math geometry questions

(1a) △PBL and △PBM are congurent with PB = PB, ∠PBL = ∠PBM and ∠PLB = ∠PMB, reason: AAS

Then we have PL = PM and BL = BM.

△PAN and △PCN are congurent with PN = PN, ∠PNA = ∠PNC and NA = NC, reason: SAS

Then we have PA = PC

△PAL and △PCM are congurent with PL = PM, PA = PC and ∠PMC = ∠PLA, reason: RHS

(b) Now,

BA + BC = BL - AL + BM + MC

Since △PAL and △PCM are congurent, AL = MC, so

BA + BC = BL + BM = 2BM

(3a) PR = QR (△PQR is equil) and RS = RT (△RST is equil)

Also ∠PRS = ∠QRT = 60

So △PRS and △QRT are congurent (SAS), giving PS = QT

(b) Since △PRS and △QRT are congurent, ∠RQT = ∠RPS, i.e. ∠SQU = ∠RPS

Also ∠PSR = ∠QSU (Vert. opp ∠s)

So △PRS and △QUS are similar (AAA)

Thus, ∠QSU = ∠PSR = 60

(6) From the given, we have ∠ABP = ∠APB and ∠CBQ = ∠CQB for AB = AP and CB = CQ resp.

Let ∠ABP = ∠APB = x and ∠CBQ = ∠CQB = y, then

∠PCB = 180 - 2y

∠PAB = 180 - 2x

∠PCB + ∠PAB = 90

360 - 2x - 2y = 90

x + y = 135

∠ABP + ∠QBC = x + y = 135

90 + 2∠PBQ = 135

∠PBQ = 22.5

(7) ∠AEF = ∠AFE

∠FAE = 180 - (x + y)

∠AEF = ∠AFE = (x + y)/2

x = z + ∠CED = z + ∠AEF = z + (x + y)/2

x = z + x/2 + y/2

x/2 = y/2 + z

x = y + 2z