# Find the area of the ΔDBC

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

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

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

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

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