# Geometry question about triangle and a circle

We have triangle $$ABC$$, and we construct a circle on side AC to become its diameter.
This circle contains the middle point of side $$BC$$ and intersects side $$AB$$ in point D, in ratio of $$AD:DB=1:2$$. If $$AB$$ is $$3cm$$ what is the area of triangle $$ABC$$?

My attempt: $$CD$$ seems to be the height of triangle(because it's at $$90$$ degrees), perpendicular to $$AC$$. I used that $$CD=\sqrt{AD*DB}$$. So I have height and base of triangle which turned out that area is $$3\sqrt{2}/2$$, but solution seems to be $$3\sqrt{2}$$. Why?

• Why do you think that $CD \perp AB$? – Dbchatto67 Apr 21 at 7:54
• Is it not that triangle $ADC$ is a right triangle? – Darko Dekan Apr 21 at 8:03
• Why? Can you explain? I can't see any such thing. – Dbchatto67 Apr 21 at 8:05
• Angle in a semicircle is always 90 degrees when one side is a diameter of a triangle. – Darko Dekan Apr 21 at 8:07
• I used Thales's theorem. – Darko Dekan Apr 21 at 8:07

Refer to the figure:

$$\hspace{5cm}$$

The angles $$ADC$$ and $$AEC$$ are right, because both subtend the diameter $$AC$$.

From $$AD:BD=1:2$$ and $$AB=3$$ we can find $$AD=1$$ and $$BD=2$$.

The line $$AE$$ is both height and median, it implies the triangle $$ABC$$ is isosceles. Hence, $$AC=3$$.

From the right triangle $$ACD$$ we can find $$CD=\sqrt{AC^2-AD^2}=2\sqrt{2}$$.

Finally, the area of the triangle $$ABC$$ is $$\frac12 \cdot AB\cdot CD=\frac12 \cdot 3\cdot 2\sqrt{2}=3\sqrt{2}.$$