Inscribed circle and altitude of triangle ?

Forum|Forum|5 years ago
December 1, 2020
2 replies
8159 views

Hello Everyone.
Given :
inscribed circle of triangle ABC touches its 3 sides at A', B', C'. (see Figure)
A'A'' is a altitude of triangle A'B'C'.
Prove :
C'B / C'A'' = B'C / B'A'' ( blue / green = red / pink )

Thanks in advance for your time and help.
Regards.
Loi.

Math Homework

ttokoro

21-Topaz I

t.t.

W

Werner_E

25-Diamond I

@ttokoro wrote:

Not equal.

No, you are wrong! The two ratios ARE equal. For the example posted approx. 2.24224.

Check your calculations or your interpretation of the position of A". Guess you misinterpreted the latter. A'A'' is perpendicular to B'C' ! The coordinates are approx. A'' (3.92964 / 2.26219).