4 coplanar points

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We put the points in a coordinate system such that

$C=(0,0) \quad D=(7,0) \quad B=(5\cdot\cos{120°},5\cdot\sin{120°})=\left(-\dfrac{5}{2},\dfrac{5\sqrt{3}}{2}\right)$

And $A$ lies on the line $r:y=x\cdot \tan{60°}=x\sqrt{3}$

We now need only to find the intersection between $r$ and $BD$

The line that passes through $BD$ is

$y = \dfrac{\dfrac{5\sqrt{3}}{2}-0}{-\dfrac{5}{2} - 7}(x-7)=-\dfrac{5\sqrt{3}}{19}(x-7)$

To find the intersection

$x\sqrt{3}=-\dfrac{5\sqrt{3}}{19}(x-7)\iff x=\dfrac{35}{24}\Longrightarrow y=\dfrac{35\sqrt{3}}{24}$

Therefore the lenght of $AC$ is

$AC=\sqrt{\left(\dfrac{35}{24}\right)^2+\left(\dfrac{35\sqrt{3}}{24}\right)^2}=\dfrac{35}{12}≈\boxed{2.92}$

Consider the triangles $BCA$ , $ACD$ and $BCD$

Now, we use the fact that if in a $\triangle XYZ$ if $XY=z, YZ=x$ and $ZX=y$ and $\angle XYZ$ $=60^{0}$ , then $y^{2}=x^{2} + z^{2}-2xz cos 60^{0}=x^{2}+z^{2}-xz$

Thus, we get that if $CA=a$ , then

$BA^{2}=5^{2}+a^{2}-5a$ and $DA^{2}=7^{2}+a^{2}-7a$

By $\triangle BCD$ , $BD^{2}=5^{2} +7^{2}-2(5)(7)cos 120^{o}=5^{2}+7^{2}+35$

Now, A lies on BD if and only if $BA+AD=BD$

i.e. $\sqrt{5^{2}+a^{2}-5a} + \sqrt{7^{2}+a^{2}-7a} =\sqrt{5^{2}+7^{2}+35}$ $\implies (5^{2}+a^{2}-5a)+(7^{2}+a^{2}-7a)+2\sqrt{(5^{2}+a^{2}-5a)(7^{2}+a^{2}-7a)}=5^{2}+7^{2}+35$

$\implies (-2a^{2}+12a+35)^{2}=4(5^{2}+a^{2}-5a)(7^{2}+a^{2}-7a)$

$\implies 4a^{4}-48a^{3}+4a^{2}+840a+1225=4a^{4}-48a^{3}+436a^{2}-1680a+4900$

$\implies -432a^{2}+2520a-3675=0$

$\implies a=\frac{35}{12}$

Thus,

$\boxed {CA=2.92}$