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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?

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

$\hspace{5cm}$![enter image description here

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 $\frac12 \cdot AB\cdot CD=\frac12 \cdot 3\cdot 2\sqrt{2}=3\sqrt{2}.$

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