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# ABC is a Δ of which ∠B is obtuse. D is the circumcentre, DF⊥ AC, DG ⊥ AB, DH ⊥BC. Prove that DG + DH = R + r + DF, where r is the?

ABC is a Δ of which ∠B is obtuse. D is the circumcentre, DF⊥ AC, DG ⊥ AB, DH ⊥BC. Prove that DG + DH = R + r + DF, where r is the in-radius and R the circum-radius of ABC.

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- IndicaLv 75 years agoFavourite answer
Using circle geometry DH=RcosA, DG=RcosC, DF=RcosB where DF<0 because B is obtuse

∴ DF+DG+DH = R(cosA+cosB+cosC)

Then use these familiar relationships (proofs are easy)

cosA+cosB+cosC = 1+4sin(A/2)sin(B/2)sin(C/2) and r = 4Rsin(A/2)sin(B/2)sin(C/2)

Hence R(cosA+cosB+cosC) = r+R which gives result

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