S1=CK.AB/2;S2=HK.AB/2
=>S1.S2=\(\dfrac{AB^2.\left(CK.HK\right)}{4}\)
=>\(\sqrt{S1.S2}=\dfrac{AB.\sqrt{CK.HK}}{2}\)
ta có góc KBH= góc KCA
=> tam giac khb dong dang tam giac akc (g.g)
hk/ak=bk/ck=>ck.hk=ak.bk
mk^2=ak.bk(theo he uoc luong tam giac)
=>mk=\(\sqrt{ck.hk}\)
=>\(\sqrt{S1.s2}=\dfrac{AB.MK}{2}=S\left(DPCM\right)\)