# Trapezoid and angle bisectors

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 $$AB=5DC$$.

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 $$DC=x\Rightarrow AB=5x$$.

• 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 $y$. Commented Apr 6, 2022 at 11:16
• @IvanKaznacheyeu, thank you for the response! How can I find the cosine of $BAO$?
– Hipo
Commented Apr 6, 2022 at 11:24
• Cosine of BAO can be found by the half angle formula. Commented Apr 6, 2022 at 11:37
• @IvanKaznacheyeu, thank you! We have $\cos\measuredangle ADC=-\cos\measuredangle BAD=-\frac23$. How do we know if $\measuredangle CDO$ is acute or obtuse?
– Hipo
Commented Apr 6, 2022 at 11:58
• $DO$ is angle bisector, then $\angle CDO=\frac{1}{2}\angle ADC < 90°$ Commented Apr 6, 2022 at 12:22

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 $$x^2-\frac23\sqrt{3}x-5=0\implies x=\frac53\sqrt{3}$$