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#293096 Cm $ab+bc+ca=p^2+r^2+4Rr$

Posted by winter_29 on 09-01-2012 - 22:30 in Hình học phẳng

Có S=abc/4R=rp =>4Rr=abc/p
r^2.p=S^2/p=(p-a)(p-b)(p-c) (hê rông)
đăng thứccần c/m
<=>p(ab+bc+ca)=p^3+(p-a)(p-b)(p-c)+abc
<=>dpcm