Math Help - Show that line passes through a point

1. Show that line passes through a point

Given is a triangle ABC. A circle with center $D$ is tangent to $BC, AB$ and $AC$ at points that lie outside the given triangle. Show that $AD$ passes through the circumcircle of triangle $BCD$.

I have no idea how to attempt this, and to begin with I don't really know the method of construction of such a circle, ie. how do I make a circle that is tangent to three given lines?

Thanks for help.

2. Originally Posted by atreyyu
Given is a triangle ABC. A circle with center $D$ is tangent to $BC, AB$ and $AC$ at points that lie outside the given triangle. Show that $AD$ passes through the circumcircle of triangle $BCD$.

I have no idea how to attempt this, and to begin with I don't really know the method of construction of such a circle, ie. how do I make a circle that is tangent to three given lines?

Thanks for help.
Hi atreyyu,

if you extend the three sides of the triangle such that the triangle sides
are now line segments of these lines, then, there are 3 circles that touch one side of the triangle tangentially
and touch the extended lines tangentially.
If you bisect the appropriate external angle you will have a line whose points are
the same perpendicular distance from the extended line and a side of the triangle.
Do the same for the other relevant extended line and you will find the centre of the circle with centre D (of which there are 3).

That can get you started geometrically.

3. Could you show this step by step? I don't really see it.

4. Let AB touches the circle at E, BC at F and AC at G. Since AE and AG are the tangents to the circle, they are equal.
AD is the angle bisector of A and DE is perpendicular to BC, hence AD is perpendicular bisector of BC. So the circumcenter of the circle passing through BCD lies on AD.

5. Hello, atreyyu!

A sketch would help . . .

Given is a triangle ABC.
A circle with center $D$ is tangent to $AB, BC, AC$
at points that lie outside the given triangle.

Show that $AD$ passes through the circumcircle of $\Delta BCD$.

This is one possible diagram . . .

Code:
                     \
* * *   \
*           *\
*               *
*                 *
\   /
*                   *\ /
*         *         * o B
*         D         */ \
/   \
*                 *     \
*               *       \
*           */         \
- - - * * * - o - - - - - o - - -
/C           A\
/               \

Circle $D$ is tangent to side $BC$
. . and tangent to sides $AB$ and $AC$ extended.

(There are two more locations for circle $D.$)

They could have said: "Circle D is externally tangent to triangle ABC."
. .

6. And could you explain step-by-step how to construct that circle with compass/straightedge?

7. The attachment is a simplistic version,
for a right-angled isosceles triangle.
Unfortunately, the labelling does not match sa-ri-ga-ma's post.

You may bisect 2 external angles,
or one of them and the angle at A.
They intersect at the same point.

Draw the 2 green circles at the vertices B and C.
Bisect those external angles with the purple circles and pair of light-blue circles
(you may prefer arcs).
The purple circles are the same size.
The light-blue circles are also of equal size.

The centres are marked.
Draw the bisectors through the points of intersection of the purple circles
and the points of intersection of the light-blue circles.

The bisectors intersect at D.

It's the same procedure for other triangles.

8. Thanks a lot, I finally drew the darn thing.
Could you give me a hand with the proof? The thing seems obvious when I look at it now, but I still don't know how to proceed.

9. If we bisect angle BAC, the centre D of the large externally tangential circle
lies on this line also.

We find the centre of a circumcircle by bisecting the sides of the triangle housing it.

This means the centre of the circumcircle of triangle BDC is on the bisector of side BC.

The centre D of the external circle also lies on this line.

This is a very simple and convenient case, however
and there is more to the general case.

Can you draw a scalene triangle and work out a proof for that,
using these posts as a guide?

10. Eh, nope. When the triangle gets scalene, my reasoning is laid waste.

11. It involves drawing a triangle with 3 sides of different lengths.
Extend the sides of the triangle out again.
Bisect angles, find the centre of the externally tangential circle.
Bisect the side of the scalene triangle that the external circle touches.
The centre of the circumcircle lies on that bisector.

12. OK, OK. I have already managed to draw a scalene triangle, etc etc. I see it that the centre of the circumcircle lies on the bisector. But what you wrote before is in no way relevant to a scalene triangle, as, forn instance, bisector of BC is not the bisector of angle at A anymore.
I really don't know what do now.

13. Originally Posted by atreyyu
OK, OK. I have already managed to draw a scalene triangle, etc etc.

there is no need to be impatient

I see it that the centre of the circumcircle lies on the bisector. But what you wrote before is in no way relevant ?? to a scalene triangle, as, for instance, bisector of BC is not the bisector of angle at A anymore.

Yes, that's true, I thought it might be helpful to leave you something to do,