Question

prove Apollo theorem ​

Answer 1

Answer:

1. SOLUTION

theorem can be proved as a special case of Stewart's theorem, or can be proved using vectors (see parallelogram law). The following is an independent proof using the law of cosines.[1]

theorem can be proved as a special case of Stewart's theorem, or can be proved using vectors (see parallelogram law). The following is an independent proof using the law of cosines.[1]Let the triangle have sides a, b, c with a median ddrawn to side a. Let m be the length of the segments of a formed by the median, so m is half of a. Let the angles formed between a and d be θ and θ′ where θ includes b and θ′ includes c. Then θ′ is the supplement of θ and cos θ′ = −cos θ. The law of cosines for θ and θ′ states

theorem can be proved as a special case of Stewart's theorem, or can be proved using vectors (see parallelogram law). The following is an independent proof using the law of cosines.[1]Let the triangle have sides a, b, c with a median ddrawn to side a. Let m be the length of the segments of a formed by the median, so m is half of a. Let the angles formed between a and d be θ and θ′ where θ includes b and θ′ includes c. Then θ′ is the supplement of θ and cos θ′ = −cos θ. The law of cosines for θ and θ′ states{\displaystyle {\begin{aligned}b^}&=m^{2}+d^{2}-2dm\cos \theta \\c^}&=m^{2}+d^{2}-2dm\cos \theta '\\&=m^{2}+d^{2}+2dm\cos \theta .\,\end{aligned}

theorem can be proved as a special case of Stewart's theorem, or can be proved using vectors (see parallelogram law). The following is an independent proof using the law of cosines.[1]Let the triangle have sides a, b, c with a median ddrawn to side a. Let m be the length of the segments of a formed by the median, so m is half of a. Let the angles formed between a and d be θ and θ′ where θ includes b and θ′ includes c. Then θ′ is the supplement of θ and cos θ′ = −cos θ. The law of cosines for θ and θ′ states{\displaystyle {\begin{aligned}b^}&=m^{2}+d^{2}-2dm\cos \theta \\c^}&=m^{2}+d^{2}-2dm\cos \theta '\\&=m^{2}+d^{2}+2dm\cos \theta .\,\end{aligned}Add these equations to obtain

theorem can be proved as a special case of Stewart's theorem, or can be proved using vectors (see parallelogram law). The following is an independent proof using the law of cosines.[1]Let the triangle have sides a, b, c with a median ddrawn to side a. Let m be the length of the segments of a formed by the median, so m is half of a. Let the angles formed between a and d be θ and θ′ where θ includes b and θ′ includes c. Then θ′ is the supplement of θ and cos θ′ = −cos θ. The law of cosines for θ and θ′ states{\displaystyle {\begin{aligned}b^}&=m^{2}+d^{2}-2dm\cos \theta \\c^}&=m^{2}+d^{2}-2dm\cos \theta '\\&=m^{2}+d^{2}+2dm\cos \theta .\,\end{aligned}Add these equations to obtain{\displaystyle b^{2}+c^{2}=2(m^{2}+d^{2})}

theorem can be proved as a special case of Stewart's theorem, or can be proved using vectors (see parallelogram law). The following is an independent proof using the law of cosines.[1]Let the triangle have sides a, b, c with a median ddrawn to side a. Let m be the length of the segments of a formed by the median, so m is half of a. Let the angles formed between a and d be θ and θ′ where θ includes b and θ′ includes c. Then θ′ is the supplement of θ and cos θ′ = −cos θ. The law of cosines for θ and θ′ states{\displaystyle {\begin{aligned}b^}&=m^{2}+d^{2}-2dm\cos \theta \\c^}&=m^{2}+d^{2}-2dm\cos \theta '\\&=m^{2}+d^{2}+2dm\cos \theta .\,\end{aligned}Add these equations to obtain{\displaystyle b^{2}+c^{2}=2(m^{2}+d^{2})}as required.

PROVED

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Answer 2

Answer:

Choose the origin of the rectangular form of the Cartesian coordinates at the point O and the x-axis coming along the sides MN and also OY as y – axis. If in case MN = 2a, then the coordinates of the points M, as well as N, are (a, 0) and (- a, 0) respectively. If coordinates of the point L are (b, c), then

LO² = (C – 0)² + (b – 0)² , (Since the coordinates of the point O are {0, 0})

= c² + b²;

LM² = (c – 0)² + (b + a) ² = c² + (a + b)²

MO² = (0 – 0)² + (- a – 0)² = a²

also, LN² = (c – 0) ² + (b – a) ² = c² + (a – b)²

Therefore, LN² + LM² = c² + (a + b) ² + c² + (b – a)²

= 2c² + 2 (a² + b²)

= 2(b² + c²) + 2a²

= 2LO² + 2MO²

= 2 (LO² + MO²).

= 2(MO² + LO²). {Hence Proved}