f.4 maths circles

17
(a)
make the line OA
∠AOB=2∠ARC
∠AOR=180-2∠ARC

So ∠ORQ=360-∠OAQ-∠AQR-∠AOR=2∠ARC
So ∠ARC=∠ARQ

(b)
∠ACR=∠AQR
AR=AR
∠ARC=∠ARQ
∆ACR≣∆ARQ (AAS)

(c)
Let the intersection of AR and CQ is E
Then ∠CER=90

∠ABC=90-∠ARC
∠QCR=90-∠ARC

So ∠ABC=∠QCR
BA//CQ

(d)
Since
∆PCQ ∼ ∆PBA
AP/PQ=BP/CP=AB/QC
9/(9+AQ)=BP/(3+BP)=5/CQ

Since AC=4cm; AQ=4cm
We have
9/(9+AQ)=5/CQ
65=9CQ
CQ=7.222cm

18
(a)
Consider ∠PDB
∠PDB=180-90-∠ODC=90-∠ACD
On the other hand
∠ABC=180-∠BAC-∠ACD=90-∠ACD

So ∠PDB=∠ABC
PB=PD

(b)
∆OAP≣∆OPD (RHS)
PA=PD
Since diameter is AC=AB
where AB=PA+PB=PA+PD
So PA+PD is equal to the diameter of the circle

2
18
(a)
Consider ∠PCD=105
Draw the line OB and OC
Then ∠OBA=∠OCD=90

We have ∠OCB=∠OBC=15
So ∠ABC=90-15=75
∠APB=55

So ∠PAB=180-75-55=50