a) Xet tu giac ABNM co: 
goc BAC= 90 do (gt) 
goc MNB= 90 do vi goc MNC= 90 do (goc noi tiep chan nua duong tron) 
=> goc BAC + goc MNB= 180 do 
=> tu giac ABNM noi tiep dc duong tron 
Xet tu giac ABCI: 
goc BAC= 90 do 
goc BIC = 90 do (goc noi tiep chan nua duong tron) 
=> A va I cung nhin BC duoi 1 goc vuong 
=>Tu giac ABCI noi tiep dc duong tron 
b)Ta co: goc ANM= goc ABM ( cung chan cung AM) 
goc INM = goc ICM ( cung chan cung IM) 
Ma goc ABM= goc ICM ( cung chan cung AI) 
=>goc ANM= goc INM 
Vay : NM là tia phân giác của góc ANI. 
c) Ta co : tam giac BNM ~ tam giac BIC(g-g) 

=>BM/BC=BN/BI 
=>BM.BI=BC.BN (1) 
mat khac: tam giac MNC ~ tam giac BAC(g-g) 
=>CM/BC=CN/CA 
=>CM.CA=BC.CN (2) 
Tu (1) va (2), suy ra: 
BM.BI +CM.CA= BC.BN+BC.CN=BC(BN+CN)=BC.BC=BC^2 
Ma BC^2=AB^2+AC^2 ( D/l Pytagore) 
Vay: BM.BI+CM.CA=AB^2+AC^2.

a) Xet tu giac ABNM co:
goc BAC= 90 do (gt)
goc MNB= 90 do vi goc MNC= 90 do (goc noi tiep chan nua duong tron)
=> goc BAC + goc MNB= 180 do
=> tu giac ABNM noi tiep dc duong tron
Xet tu giac ABCI:
goc BAC= 90 do
goc BIC = 90 do (goc noi tiep chan nua duong tron)
=> A va I cung nhin BC duoi 1 goc vuong
=>Tu giac ABCI noi tiep dc duong tron
b)Ta co: goc ANM= goc ABM ( cung chan cung AM)
goc INM = goc ICM ( cung chan cung IM)
Ma goc ABM= goc ICM ( cung chan cung AI)
=>goc ANM= goc INM
Vay : NM là tia phân giác của góc ANI.
c) Ta co : tam giac BNM ~ tam giac BIC(g-g)
=>BM/BC=BN/BI
=>BM.BI=BC.BN (1)
mat khac: tam giac MNC ~ tam giac BAC(g-g)
=>CM/BC=CN/CA
=>CM.CA=BC.CN (2)
Tu (1) va (2), suy ra:
BM.BI +CM.CA= BC.BN+BC.CN=BC(BN+CN)=BC.BC=BC^2
Ma BC^2=AB^2+AC^2 ( D/l Pytagore)
Vay: BM.BI+CM.CA=AB^2+AC^2.

a) Xet tu giac ABNM co:
goc BAC= 90 do (gt)
goc MNB= 90 do vi goc MNC= 90 do (goc noi tiep chan nua duong tron)
=> goc BAC + goc MNB= 180 do
=> tu giac ABNM noi tiep dc duong tron
Xet tu giac ABCI:
goc BAC= 90 do
goc BIC = 90 do (goc noi tiep chan nua duong tron)
=> A va I cung nhin BC duoi 1 goc vuong
=>Tu giac ABCI noi tiep dc duong tron
b) Ta co: goc ANM= goc ABM ( cung chan cung AM)
goc INM = goc ICM ( cung chan cung IM)
Ma goc ABM= goc ICM ( cung chan cung AI)
=>goc ANM= goc INM
Vay : NM là tia phân giác của góc ANI.
c) Ta co : tam giac BNM ~ tam giac BIC(g-g)
=>BM/BC=BN/BI
=>BM.BI=BC.BN (1)
mat khac: tam giac MNC ~ tam giac BAC(g-g)
=>CM/BC=CN/CA
=>CM.CA=BC.CN (2)
Tu (1) va (2), suy ra:
BM.BI +CM.CA= BC.BN+BC.CN=BC(BN+CN)=BC.BC=BC^2
Ma BC^2=AB^2+AC^2 ( D/l Pytagore)
Vay: BM.BI+CM.CA=AB^2+AC^2.