c) Áp dụng HTL trong tam giác vuông ABO có: AB^2 = BE. BO
Áp dụng HTL trong tam giác vuông ABC có AB^2 =BH.BC
=> BE.BO = BH.BC
\(\eqalign{
& BE.BO = BH.BC \Rightarrow {{BE} \over {BH}} = {{BC} \over {BO}} \cr
& Xet\,\,\Delta BHE\,\,va\,\,\,\Delta BOC\,\,co: \cr
& \angle B\,\,chung; \cr
& {{BE} \over {BH}} = {{BC} \over {BO}}\,\,\left( {cmt} \right); \cr
& \Rightarrow \Delta BHE \sim \Delta BOC\,\,\left( {c.g.c} \right) \cr
& \Rightarrow \angle BEC = \angle BCA\,\,\left( {2\,\,goc\,\,tuong\,\,ung} \right). \cr
& Ma\,\,\angle BCA = \angle BAH\,\,\left( {cung\,\,phu\,\,voi\,\,\angle HAC} \right) \cr
& \Rightarrow \angle BEC = \angle BAH \cr} \)
\(\eqalign{
& d)\,\,Xet\,\,\Delta ABH\,\,va\,\,\Delta CBA\,\,co: \cr
& \angle B\,\,chung \cr
& \angle BAH\, = \angle BCA\,\,\left( {cung\,\,phu\,\,voi\,\,\angle HAC} \right) \cr
& \Rightarrow \Delta ABH \sim \Delta CBA\,\,\left( {g.g} \right) \cr
& \Rightarrow {{AB} \over {AH}} = {{CB} \over {CA}} \cr
& \Rightarrow {{AB} \over {2AI}} = {{CB} \over {2CO}} \Rightarrow {{AB} \over {AI}} = {{CB} \over {CO}} \cr
& Xet\,\,\Delta ABI\,\,va\,\,\,\Delta CBO\,\,co: \cr
& \angle BAI = \angle BCO\,\,\,\,\left( {cung\,\,phu\,\,voi\,\,\angle HAC} \right) \cr
& {{AB} \over {AI}} = {{CB} \over {CO}}\,\,\left( {cmt} \right) \cr
& \Rightarrow \,\Delta ABI \sim \Delta CBO\,\,\left( {c.g.c} \right) \cr
& \Rightarrow \angle ABI = \angle CBO\,\,\left( {dpcm} \right) \cr}\)
