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Let the in circle be P and the excircle be QLet DAC touch circle P at X & circle Q at YLet DB touch circle P at U and circle Q at VLet AB = BC = a, AX = b, CY = c and AE = EC = dLet the bisector of < BAE meet BE at TET = hd / (a + d) ......(1)Now PXA, TEA & QYC are similar trianglesSo b / r1 = c / r2 = ET / d = h / (a + d) = (b + c) / ( r1 + r2) = (a - d) / (r1 + r2) from (1)Hence (r1 + r2)h = a^2 - d^2 = h^2Therefore r1 + r2 = hSumith PeirisMoratuwaSri Lanka
Let the in circle be P and the excircle be Q
ReplyDeleteLet DAC touch circle P at X & circle Q at Y
Let DB touch circle P at U and circle Q at V
Let AB = BC = a, AX = b, CY = c and
AE = EC = d
Let the bisector of < BAE meet BE at T
ET = hd / (a + d) ......(1)
Now PXA, TEA & QYC are similar triangles
So b / r1 = c / r2 = ET / d = h / (a + d)
= (b + c) / ( r1 + r2) = (a - d) / (r1 + r2) from (1)
Hence (r1 + r2)h = a^2 - d^2 = h^2
Therefore r1 + r2 = h
Sumith Peiris
Moratuwa
Sri Lanka