Point $B$ lies on the segment $AC$. A line passing through point $A$ touches a circle with diameter $BC$ at point $M$ and intersects the circle with diameter $AB$ at point $K$. The line $MB$ intersects the circle with the diameter $AB$ at point $D$. $AK = 3$, $MK = 12$.

Find the area of the $ΔDBC$

Edit: There was just a miscalculation, I corrected my solution, now the answer is right.

My thoughts: $\angle OMK=90^{\circ }$, $\angle BKA=90^{\circ }$$OM∥BK⇒$ $\triangle ABK\sim \triangle AOM$, $OB=R$, $AB=2r$, $\frac{3}{\:15}=\frac{2r}{R+2r}⇒$ $R=8r$.

$AM^2=(2R+2r)\cdot 2r⇒ 15^2=36r^2 ⇒ r= \frac{5}{2}, R=20$.

$\triangle DBA\sim \triangle BCM$ because $AD∥CM, \frac{DB}{BM}=\frac{2r}{2R}⇒BM=8DB, DM=9DB.$

$AM\cdot MK=DM\cdot BM⇒ 15\cdot 12=9DB\cdot 8DB ⇒ DB=\sqrt{\frac{5}{2}}$

$CM^2=(2R)^2-BM^2 ⇒ CM= \sqrt{4\cdot (20)^2-\left(8^2\cdot \frac{5}{2}\right)}=\sqrt{1440}=12\sqrt{10} $ .

$S_{\triangle DBC}= \frac{CM\cdot DB}{2}=\frac{12\sqrt{10}\cdot \sqrt{\frac{5}{2}}}{2}=\frac{12\sqrt{5}\cdot\sqrt{2}\cdot \sqrt{5}}{2\sqrt{2}}=\frac{12\cdot 5}{2}=30$.

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  • 2
    $\begingroup$ How do you get $r = \frac{5}{12}$? Isn't $r^2 = \frac{15^2}{6^2}$? $\endgroup$ – rogerl Feb 19 at 22:20
  • $\begingroup$ @rogerl Thank you, so it was just a miscalculation! $\endgroup$ – Enc_23 Feb 20 at 0:19

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