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The angle bisectors of $\measuredangle BAD$ and $\measuredangle ADC$ of the trapezoid $ABCD$ $(AB\parallel CD)$ intersect at $O$. Find the lengths of $AD$ and $DC$ if $\cos\measuredangle BAD=\dfrac23,OC=\sqrt7,OB=3\sqrt{15}$ and $AB=5DC$.

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The only thing I was able to gather is that $\measuredangle AOD=90^\circ$ as $$\measuredangle DAO+\measuredangle ADO=\dfrac12\measuredangle BAD+\dfrac12\measuredangle ADC=\dfrac12(\measuredangle BAD+\measuredangle ADC)=\dfrac12 180^\circ=90^\circ\\\Rightarrow \measuredangle AOD=90^\circ$$ I don't see how I can use the given lengths as for example in triangle $BOC$ we have only 2 elements. And of course we can say $DC=x\Rightarrow AB=5x$.

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    $\begingroup$ Find cosines of $BAO$, $CDO$. Mark $AD=y$, $DC=x$, $AB=5x$. Express $DO$ in terms of $y$. Write cosine rule for $CO$ and $BO$. Then solve system of two equations for $x$ and $y$. $\endgroup$ Commented Apr 6, 2022 at 11:16
  • $\begingroup$ @IvanKaznacheyeu, thank you for the response! How can I find the cosine of $BAO$? $\endgroup$
    – Hipo
    Commented Apr 6, 2022 at 11:24
  • $\begingroup$ Cosine of BAO can be found by the half angle formula. $\endgroup$ Commented Apr 6, 2022 at 11:37
  • $\begingroup$ @IvanKaznacheyeu, thank you! We have $\cos\measuredangle ADC=-\cos\measuredangle BAD=-\frac23$. How do we know if $\measuredangle CDO$ is acute or obtuse? $\endgroup$
    – Hipo
    Commented Apr 6, 2022 at 11:58
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    $\begingroup$ $DO$ is angle bisector, then $\angle CDO=\frac{1}{2}\angle ADC < 90°$ $\endgroup$ Commented Apr 6, 2022 at 12:22

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You have already established that $\angle AOD=90^o$. Let $DC=x, AB=5x$ and $AD=y$.

If $\angle OAD=\theta$. then $\cos2\theta=\frac23\implies2\cos^2\theta-1=\frac23$ $$\implies\cos\theta=\sqrt{\frac56}\implies \cos(90-\theta)=\frac{1}{\sqrt{6}}$$

Therefore, $OD=\frac{y}{\sqrt{6}}$ and $AO=y\sqrt{\frac56}$

Now apply the Cosine Rule in triangles $ODC$ and $OAB$ giving $$7=\frac{y^2}{6}+x^2-\frac13 xy$$ and$$135=25x^2+\frac56y^2-\frac{25}{3}xy$$

Multiplying the first of these equations by $25$ and subtracting gives $$40=\frac{10}{3}y^2\implies y=2\sqrt{3}$$ Substituting this back into the first equation gives $$x^2-\frac23\sqrt{3}x-5=0\implies x=\frac53\sqrt{3}$$

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