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Proof 1 As shown in Figure 2, connect IC, IQ, PA, PB
P C, draw AD ⊥ BP at point D, IS ⊥ BQ at point S.
Q four points are concyclic, so
∠BAD = ∠PAD = ∠BCP,
∠LBIS = ∠QIS,
∠ABI = ∠CBI = ∠BCI = ∠ACI,
∠PIQ = ∠PBQ
Then ∠BAD + ∠ABI + ∠IBC + ∠CBP = 90°,
∠SIR = ∠ICP.
Also, ∠BIP = ∠RIB, thus,
IB is the tangent of the circumcircle of △BPR
IB
Also, ∠CIP = ∠RIC, hence
IC is the tangent of the circumcircle of △CPR
∠IRC = ∠ICP = ∠SIR
IS // CR.
Therefore, BR ⊥ CR.
Proof 2 Using circle ⊙I as the base circle for inversion, the image of the circle
is the line BQ, and the image of point P is R.
According to the angle-preserving property of inversion,
∠LIBP = ∠BRI, ∠ICP = ∠CRI,
= ∠IBP + ∠ICP
= 2π - ∠BIC - ∠BPC
= 2π - (π/2 + 1/2∠BAC) - (π - 1/2∠BAC)
= π
Therefore, BR ⊥ CR.
Proof 3 As shown in Figure 3, let the center of circle Γ be O, draw OD ⊥ AB at point D, OE ⊥ AI at point E,
Since AI ⊥ BC, then OE // BC.
= 90 - ∠BIE + ∠BIO
= 90
= ∠IBC + ∠OBI = ∠OBD.
Also, OB = OI, ∠ODB = ∠OEI = 90°, then
△ODB ≅ △OEI ⇒ OE = BD.
By the properties of intersecting circles, B and Q are symmetric about line O
and B and P are symmetric about line OA.
Therefore, BQ ⊥ IO.
Also, BC ⊥ IE, hence
∠CBQ = ∠OIE = ∠BOD
Note that,
∠IRB = 180°
= ∠AOB - ∠BOD - ∠IBC
= ∠AOD - 1/2∠ABC
= 90°
Draw CR' ⊥ BQ at point R'
In △BIR, by the Law of Sines,
sin ∠BIR = BI / sin ∠IRB
BI = BC / (2 cos ∠IBC)
Therefore, BR = BR'
⇔ sin ∠BIR
= 2 sin ∠IRB · cos ∠IBC · cos ∠CBQ
* sin ∠AOB
= 2 cos (∠BAO + 1/2∠ABC).
cos 1/2∠ABC · cos ∠BOD
⇔ sin ∠BAO · cos ∠ABO +
sin ∠ABO · cos ∠BA
= (cos ∠BAO + cos (∠BAO + ∠ABC)).
sin ∠AB
⇔ sin ∠BAO · cos ∠ABO
= cos ∠AFC · sin ∠ABO
⇔ sin ∠BAO = cos ∠AOE · tan ∠ABO
⇔ DO / OA = OE / AO · DO / DB ⇔ OE = BD.
The above equation is obviously true.
Proof 4 As shown in Figure 4, let the circle determined by points C, I, P
Γ1, intersecting ⊙I at point S (if the two circles are tangent, consider point C
and S coinciding, and line CS as the common tangent of the two circles).
Applying the Radical Axis Theorem to circles Γ1, ⊙I, and Γ, we have CS,
= ∠BIP + ∠CIP = ∠BIC.
Similarly, ∠RPS = ∠IPC, thus,
∠QPS = ∠QPR + ∠RPS
Note that, ∠BIC = 90° + 1/2∠A,
Then ∠QR
= 360° - (∠PQR + ∠PSR) - ∠QPS
= 360° - (90° + 1/2∠A) - (180° - 1/2∠A)
= 90°.
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