Ta có: ΔABCΔABC vuông cân tại A

⇒\hept⎧⎪⎨⎪⎩ˆBAC=900AB=ACˆABC=ˆACB=450⇒\hept{BAC^=900AB=ACABC^=ACB^=450

Lại có: \hept{ˆBAH+ˆHAC=900ˆKCA+ˆHAC=900\hept{BAH^+HAC^=900KCA^+HAC^=900

⇒ˆBAH=ˆKCA⇒BAH^=KCA^

Xét ΔABHΔABH và ΔCAK:ΔCAK:

\hept⎧⎪ ⎪⎨⎪ ⎪⎩ˆAHB=ˆCKA=900AB=AC(cmt)ˆBAH=ˆKCA(cmt)\hept{AHB^=CKA^=900AB=AC(cmt)BAH^=KCA^(cmt)

⇒ΔABH=ΔCAK(ch+gn)⇒ΔABH=ΔCAK(ch+gn)

⇒AH=CK⇒AH=CK

Có: \hept{AM⊥MBˆABM=450\hept{AM⊥MBABM^=450

 ⇒ˆMAB=450=ˆACM⇒MAB^=450=ACM^

⇒ˆBAH−ˆBAM=ˆKCA−ˆACM⇒BAH^−BAM^=KCA^−ACM^

⇒ˆHAM=ˆKCM⇒HAM^=KCM^

Ta lại có: \hept{AM⊥MCˆAMC=450\hept{AM⊥MCAMC^=450

⇒ˆMAC=450⇒MAC^=450

⇒ΔAMC⇒ΔAMC vuông cân.⇒MA=MC⇒MA=MC

Xét ΔAMHΔAMH và ΔCMK:ΔCMK:

\hept⎧⎪⎨⎪⎩AH=KC(cmt)ˆHAM=ˆKCM(cmt)AM=CM(cmt)\hept{AH=KC(cmt)HAM^=KCM^(cmt)AM=CM(cmt)

⇒ΔAMH=ΔCMK(c.g.c)⇒ΔAMH=ΔCMK(c.g.c)

⇒MK=MH.