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Since ΔABC and ΔHBC have the same circumradius, a₁ = 2R cosH = −2R cosAr = 4R sin(A/2) sin(B/2) sin(C/2)r₁ = 4R sin(A/2) cos(B/2) cos(C/2)r₁ − r = 4R sin(A/2) cos(B/2 + C/2) = 4R sin²(A/2) = 2R(1−cosA) = 2R + a₁a₁ + r₁ − r = 2Rr₁ − a₁ = 2R + r
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Since ΔABC and ΔHBC have the same circumradius,
ReplyDeletea₁ = 2R cosH = −2R cosA
r = 4R sin(A/2) sin(B/2) sin(C/2)
r₁ = 4R sin(A/2) cos(B/2) cos(C/2)
r₁ − r = 4R sin(A/2) cos(B/2 + C/2)
= 4R sin²(A/2) = 2R(1−cosA) = 2R + a₁
a₁ + r₁ − r = 2R
r₁ − a₁ = 2R + r