Question 1210147: In \triangle ABC, we have AB = AC = 4 and \angle BAC = 45^\circ. If M is the midpoint of BC, then find AM^2. Found 2 solutions by CPhill, ikleyn:Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! Let $\triangle ABC$ be a triangle with $AB = AC = 4$ and $\angle BAC = 45^\circ$. Let $M$ be the midpoint of $BC$. We want to find $AM^2$.
We can use the Law of Cosines to find $BC$.
$$BC^2 = AB^2 + AC^2 - 2(AB)(AC)\cos(\angle BAC)$$
$$BC^2 = 4^2 + 4^2 - 2(4)(4)\cos(45^\circ)$$
$$BC^2 = 16 + 16 - 32\left(\frac{\sqrt{2}}{2}\right)$$
$$BC^2 = 32 - 16\sqrt{2}$$
$$BC = \sqrt{32 - 16\sqrt{2}} = 4\sqrt{2 - \sqrt{2}}$$
Since $M$ is the midpoint of $BC$, we can use Apollonius' Theorem:
$$AB^2 + AC^2 = 2(AM^2 + BM^2)$$
$$4^2 + 4^2 = 2(AM^2 + BM^2)$$
$$32 = 2(AM^2 + BM^2)$$
$$16 = AM^2 + BM^2$$
We know $BM = \frac{1}{2}BC$, so $BM^2 = \frac{1}{4}BC^2 = \frac{1}{4}(32 - 16\sqrt{2}) = 8 - 4\sqrt{2}$.
Substituting $BM^2$ into the equation from Apollonius' Theorem:
$$16 = AM^2 + (8 - 4\sqrt{2})$$
$$AM^2 = 16 - (8 - 4\sqrt{2})$$
$$AM^2 = 8 + 4\sqrt{2}$$
Thus, $AM^2 = 8 + 4\sqrt{2}$.
Final Answer: The final answer is $\boxed{8 + 4 \sqrt{2}}$
You can put this solution on YOUR website! .
In triangle ABC, we have AB = AC = 4 and angle BAC = 45°. If M is the midpoint of BC, then find AM^2.
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This problem is simple, and I will provide a simple solution without referring to
Apollonius' Theorem, to which average student is unfamiliar 129%.
Find BC^2 using the cosine law
BC^2 = = = . (1)
The area of triangle ABC is
area = = = . (2)
Another way to calculate the area of triangle ABC is
area = . (3)
From (2), we can write
area^2 = = 32.
From (3), we can write
area^2 = .
It gives us this equation
= 32,
= = use (1) = = = = = .
ANSWER. = .
