Can Some on give me a proof of apollonius theorem?
I saw a clear explanation with example for apollonius theorem in easycalculation.Where they have explained with step by step solution.
- LearnerLv 77 years agoFavorite Answer
i) Apollonius Theorem:
"In a triangle ABC, if AD is the median to BC, then AB² + AC² = 2(AD² + BD²)"
ii) As I am not sure of your knowledge on Cosine Law of triangle, let me provide the solution using basic geometrical concepts applied in high school geometry.
iii) Draw an altitude from vertex A to BC, to meet BC at E. [E may be within B & D or withing D & C, depends upon AB < AC or AB > AC; let us here take AB < AC, so E is between B & D]
iv) Applying Pythagoras theorem to respective right triangles,
From right triangle, AEB, AB² = AE² + BE² ------ (1)
From right triangle, AED, AD² = AE² + ED² ------ (2)
(1) - (2): AB² - AD² = BE² - ED² = (BD - DE)² - ED² = BD² - 2BD*DE + ED² - ED²
==> AB² - AD² = BD² - 2BD*DE ------- (3)
v) Similarly considering the right triangles, AEC & AED, we get,
AC² - AD² = DE² - ED² = (CD + DE)² - DE² = CD² + 2CD*DE + DE² - DE²
==> AC² - AD² = CD² + 2CD*DE = BD² + 2BD*DE ------ (4)
[Since AD is the median, D is mid point of BC; hence BD = DC]
vi) Adding (3) & (4)
==> AB² + AC² - 2AD² = 2BD²
==> AB² + AC² = 2(AD² + BD²) [Proved]
- 7 years ago
I know two ways of proving. The first is based on cosine theorem. Choose any angle of triangle and express side and mediana using cosine theorem. Then equal cosines:
m^2=AB^2+(AC/2)^2 - 2AB*AC/2*cosA
BC^2=AB^2+AC^2 - 2*AB*AC*cosA
2)Using the theorem about parallelogram. (a^2+b^2)=d1^2+d2^2, where d1 and d2 are diagonals of par.
If you continue mediana until it's twice of its length and connect the end of mediana with two other vertexes you'll get a parallelogramm. One of its diagonals equals 2m.
- Will HLv 77 years ago