10. If AC and BD are the diagonals of a parallelogram ABCD show that:
a. [tex]AC ^2+ BD ^2=2\left( AB ^2+ AD ^2\right)[/tex].
b. [tex]AC ^2- BD ^2=4 \overrightarrow{ AB } \cdot \overrightarrow{ AD }[/tex].
11. In any triangle ABC , prove by vector method:
a. [tex]b=c \cos A+a \cos C[/tex]
b. [tex]a=c \cos B+b \cos C[/tex]
c. [tex]a^2=b^2+c^2-2 b c \cos A[/tex]
d. [tex]b^2=c^2+a^2-2 c a \cos B[/tex].

Express the diagonals of the parallelogram as vectors in terms of its sides.
Calculate A C 2 + B D 2 using the dot product and simplify to show it equals 2 ( A B 2 + A D 2 ) .
Calculate A C 2 − B D 2 using the dot product and simplify to show it equals 4 A B ⋅ A D .
Express the sides of the triangle as vectors and use the vector addition rule to prove the given trigonometric identities, including the cosine rule. P ro v e n ​

Explanation

Express diagonals as vectors. Let A BC D be a parallelogram. Then A C = A B + A D and B D = A D − A B .

Calculate A C 2 + B D 2 .
a. We want to show that A C 2 + B D 2 = 2 ( A B 2 + A D 2 ) .
We have


A C 2 = A C ⋅ A C = ( A B + A D ) ⋅ ( A B + A D ) = A B 2 + 2 A B ⋅ A D + A D 2
B D 2 = B D ⋅ B D = ( A D − A B ) ⋅ ( A D − A B ) = A D 2 − 2 A B ⋅ A D + A B 2
Adding these, we get
A C 2 + B D 2 = A B 2 + 2 A B ⋅ A D + A D 2 + A D 2 − 2 A B ⋅ A D + A B 2 = 2 ( A B 2 + A D 2 ) .

Calculate A C 2 − B D 2 .
b. We want to show that A C 2 − B D 2 = 4 A B ⋅ A D .
We have

A C 2 = A C ⋅ A C = ( A B + A D ) ⋅ ( A B + A D ) = A B 2 + 2 A B ⋅ A D + A D 2
B D 2 = B D ⋅ B D = ( A D − A B ) ⋅ ( A D − A B ) = A D 2 − 2 A B ⋅ A D + A B 2
Subtracting these, we get
A C 2 − B D 2 = A B 2 + 2 A B ⋅ A D + A D 2 − ( A D 2 − 2 A B ⋅ A D + A B 2 ) = 4 A B ⋅ A D .

Express triangle sides as vectors.

Let A BC be a triangle with sides a , b , c opposite to angles A , B , C respectively. Let BC = a , C A = b , A B = c . Then a + b + c = 0 .

Prove b = c cos A + a cos C .
a. We want to show that b = c cos A + a cos C .
Since a + b + c = 0 , we have b = − ( a + c ) . Taking the dot product with b , we get


b ⋅ b = − ( a + c ) ⋅ b = − a ⋅ b − c ⋅ b
b 2 = − ab cos ( π − C ) − c b cos ( π − A ) = ab cos C + c b cos A
Dividing by b , we get b = a cos C + c cos A .

Prove a = c cos B + b cos C .
b. We want to show that a = c cos B + b cos C .
Since a + b + c = 0 , we have a = − ( b + c ) . Taking the dot product with a , we get

a ⋅ a = − ( b + c ) ⋅ a = − b ⋅ a − c ⋅ a
a 2 = − ba cos ( π − C ) − c a cos ( π − B ) = ba cos C + c a cos B
Dividing by a , we get a = b cos C + c cos B .

Prove a 2 = b 2 + c 2 − 2 b c cos A .
c. We want to show that a 2 = b 2 + c 2 − 2 b c cos A .
Since a + b + c = 0 , we have a = − ( b + c ) . Taking the dot product with a , we get

a ⋅ a = ( b + c ) ⋅ ( b + c ) = b ⋅ b + 2 b ⋅ c + c ⋅ c
a 2 = b 2 + c 2 + 2 b c cos ( π − A ) = b 2 + c 2 − 2 b c cos A .

Prove b 2 = c 2 + a 2 − 2 c a cos B .
d. We want to show that b 2 = c 2 + a 2 − 2 c a cos B .
Since a + b + c = 0 , we have b = − ( a + c ) . Taking the dot product with b , we get

b ⋅ b = ( a + c ) ⋅ ( a + c ) = a ⋅ a + 2 a ⋅ c + c ⋅ c
b 2 = a 2 + c 2 + 2 a c cos ( π − B ) = a 2 + c 2 − 2 a c cos B .
Examples
These vector identities are fundamental in physics and engineering. For example, the parallelogram identity relates the lengths of the sides and diagonals of a parallelogram, which can be used to analyze forces acting on an object. The cosine rule is used extensively in navigation and surveying to calculate distances and angles.

Answered by GinnyAnswer | 2025-07-03

We proved the properties of the diagonals of a parallelogram and several relationships in a triangle using vector methods. The proofs rely on vector addition and the properties of dot products. The results demonstrate fundamental relationships in geometry using vectors.
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Answered by Anonymous | 2025-07-04