Can every right triangle be inscribed in a circle?
If so, we could infer: ": in any right triangle, the segment which cuts the hypotenuse in two and joins the opposite angle is equal to half the hypotenuse". (as $r=\frac{1}{2d}$).
Thanks!
[Math] Can any right triangle be inscribed in a circle
circlesgeometrytriangles
Related Solutions
Let $O$ be the centre of the incircle. Join $O$ to the three vertices.
It is not absolutely clear what ratio $12:5$ means. Let $M$ be the middle of the base. Maybe (i) $BO:OM=12:5$, or maybe (ii) $BO:OM=5:12$. (It will turn out that (ii) can't happen.)
We examine the consequences of (i). Let the base be $2y$ (I don't like fractions). Let $BO=12t$ and let $OM=5t$.
We calculate the area of triangle $BAC$ in two different ways. The area is half of base times height, so it is $(1/2)(2y)(17t)$, that is, $17yt$.
The area is also the sum of the areas of $\triangle BOA$, $\triangle BOC$, and $\triangle OAC$. This sum is $(1/2)(30)(5t)+(1/2)(30)(5t)+(1/2)(2y)(5t)$, which is $150t +5yt$. We conclude that $$17yt=150t+5yt.$$ Thus $12y=150$, and therefore $2y=25$.
Next we examine possibility (ii). The analysis is similar. We find pretty quickly that $y$ is "too big" (the Triangle Inequality is violated).
In the diagram above $A$ is the centre of the circle and $CB$ is thus the diameter. Point $D$ is an arbitrary point in the circumference.
In $\triangle ACD$ $\angle CDA = \alpha$ since $AC = AD$
Thus
$$2 \cdot \alpha + \theta = 180 ^\circ$$
In $\triangle ADB$ $\angle ADB = \beta$ since $AB = AD$
Thus
$$2 \cdot \beta + (180^\circ - \theta) = 180 ^\circ$$
Add both these together
$$2 \cdot \alpha + 2 \cdot \beta + 180 ^\circ = 360 ^\circ$$
Which we can easily rearrange to show:
$$ \alpha + \beta = 90 ^\circ $$
We have thus proved $\angle CDB$ is a right angle as required.
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Best Answer
You are mixing up two different, unrelated questions.
Any triangle can be inscribed in a circle.
Right triangles are the only ones where the circumcenter lies on one of the sides.