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Level: Mathematics Education, High School, Honors Geometry, College.
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Monday, March 2, 2020
Geometry Problem 1456: Altitudes, Orthic Triangle, Circumcircle, Parallel, Similarity, Area
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<BB1C1 = < BCC1 = 90 - B = <BAA1 = <BH(B)H(C), hence H(B)H(C) // B1C1 and so on for the other sides of the 2 Tr.s A1B1C1 and H(A)H(B)H(C)
ReplyDelete90 - A = <ABH(B) = <ACC1 = <ABC1 so Tr.s BC1H(c) & BHH(C) are congruent ASA
Hence H(C) is the midpoint of HC1 and similarly H(B) is the mid point of HB1
From the Mid Point theorem, H(B)H(C) = B1C1/2
And therefore S(A1B1C1) = 4.S(H(A)H(B)H(C))
Sumith Peiris
Moratuwa
Sri Lanka