a) Ta có:

{ˆBchungˆADB=ˆCFB=900⇒ΔADB∼ΔCFB(g.g)⇒BABD=BCBF⇒BF.BA=BD.BC{B^chungADB^=CFB^=900⇒ΔADB∼ΔCFB(g.g)⇒BABD=BCBF⇒BF.BA=BD.BC

Lại có:

⎧⎪⎨⎪⎩BABD=BCBFˆBchung⇒ΔBFD∼ΔBCA(c.g.c)⇒ˆBFD=ˆBCA{BABD=BCBFB^chung⇒ΔBFD∼ΔBCA(c.g.c)⇒BFD^=BCA^

b) Ta có:

{ˆHFB=ˆHEC=900ˆFHB=ˆEHC(dd)⇒ΔFHB∼ΔEHC(g.g)⇒HFHE=HBHC⇒HF.HC=HE.HB{HFB^=HEC^=900FHB^=EHC^(dd)⇒ΔFHB∼ΔEHC(g.g)⇒HFHE=HBHC⇒HF.HC=HE.HB

Lại có:

HFHE=HBHC⇒HFHB=HEHCHFHE=HBHC⇒HFHB=HEHC

Khi đó:

⎧⎪⎨⎪⎩ˆEHF=ˆCHB(dd)HFHB=HEHC⇒ΔEHF∼ΔCHB(g.g)⇒ˆFEH=ˆBCH⇒ˆFEB=ˆFCB{EHF^=CHB^(dd)HFHB=HEHC⇒ΔEHF∼ΔCHB(g.g)⇒FEH^=BCH^⇒FEB^=FCB^

c) Ta có:

{ˆCchungˆCDH=ˆCFB=900⇒ΔCDH∼ΔCFB(g.g)⇒CDCF=CHCB⇒CF.CH=CD.CB⇒CF.CH+BF.BA=CD.CB+BD.BC=BC.(CD+BD)⇒CF.CH+BF.BA=BC2{C^chungCDH^=CFB^=900⇒ΔCDH∼ΔCFB(g.g)⇒CDCF=CHCB⇒CF.CH=CD.CB⇒CF.CH+BF.BA=CD.CB+BD.BC=BC.(CD+BD)⇒CF.CH+BF.BA=BC2

d) Ta có:

{ˆAchungˆAEB=ˆAFC=900⇒ΔAEB∼ΔAFC(g.g)⇒AEAF=ABAC⇒AEAB=AFAC{A^chungAEB^=AFC^=900⇒ΔAEB∼ΔAFC(g.g)⇒AEAF=ABAC⇒AEAB=AFAC

Khi đó:

⎧⎪⎨⎪⎩ˆAchungAEAB=AFAC⇒ΔAEF∼ΔABC(g.g)⇒ˆAFE=ˆACB⇒ˆIFB=ˆACB⇒ˆIFB=ˆICE{A^chungAEAB=AFAC⇒ΔAEF∼ΔABC(g.g)⇒AFE^=ACB^⇒IFB^=ACB^⇒IFB^=ICE^

Như vậy:

{ˆIchungˆIFB=ˆICE⇒ΔIFB∼ΔICE(g.g)⇒IFIC=IBIE⇒IB.IC=IF.IE(∗){I^chungIFB^=ICE^⇒ΔIFB∼ΔICE(g.g)⇒IFIC=IBIE⇒IB.IC=IF.IE(∗)

Lại có:

ˆIFD=ˆIFB+ˆBFD=ˆICE+ˆBCA=2ˆBCA(1)IFD^=IFB^+BFD^=ICE^+BCA^=2BCA^(1)

Mà ΔBEC;ˆE=900;OΔBEC;E^=900;O là trung điểm của BCBC

⇒OB=OC=OE=12BC⇒OB=OC=OE=12BC

⇒ΔCOE⇒ΔCOE cân ở OO

⇒ˆBOE=2ˆOCE⇒ˆIOE=2ˆBCA(2)⇒BOE^=2OCE^⇒IOE^=2BCA^(2)

Từ (1),(2)⇒ˆIFD=ˆIOE(1),(2)⇒IFD^=IOE^

Khi đó:

{ˆIchungˆIFD=ˆIOE⇒ΔIFD∼ΔIOE(g.g)⇒IFIO=IDIE⇒IO.ID=IF.IE(∗∗){I^chungIFD^=IOE^⇒ΔIFD∼ΔIOE(g.g)⇒IFIO=IDIE⇒IO.ID=IF.IE(∗∗)

Từ (∗),(∗∗)⇒IO.ID=IB.IC