Question 1210147
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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%.



<pre>
Find BC^2 using the cosine law

    BC^2 = {{{4^2 + 4^2 - 2*4*4*cos(45^o)}}} = {{{32 - 32*(sqrt(2)/2)}}} = {{{32-16*sqrt(2)}}}.    (1)


The area of triangle ABC is

    area = {{{(1/2)*AB*AC*sin(45^o)}}} = {{{(1/2)*4*4*(sqrt(2)/2)}}} = {{{4*sqrt(2)}}}.    (2)


Another way to calculate the area of triangle ABC is

    area = {{{(1/2)*BC*AM}}}.   (3)


From (2), we can write

    area^2 = {{{(4*sqrt(2))^2}}} = 32.


From (3), we can write

    area^2 = {{{(1/4)*BC^2*AM^2}}}.


It gives us this equation

    {{{(1/4)*BC^2*AM^2}}} = 32,

    {{{AM^2}}} = {{{(32*4)/(BC^2)}}} = use (1) = {{{128/(32-16*sqrt(2))}}} = {{{8/(2-sqrt(2))}}} = {{{(8*(2+sqrt(2)))/(4-2)}}} = {{{4*(2+sqrt(2))}}} = {{{8+4sqrt(2)}}}.


<U>ANSWER</U>.  {{{AM^2}}} = {{{8 + 4*sqrt(2)}}}.
</pre>

Solved adequately in a simple way.