hi Agnishom

I think I'd use Euclidean geometry and specifically congruent triangles for the first two.  And vector geometry for the last.

Still thinking about the altitudes.

edit:  added a diagram for the inscribed centre.

BD and CD are bisectors that meet at D

Make perpendiculars DE and DF.

Prove DE = DF by congruency.

Then make the last perpendicular and show, again by congruency, that DA is another bisector.

Bob

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Just edited above to show the first in outline.  I should think you can fill in the details.

I've got to log out soon, so I'll come back to the others later (if you haven't figured it out for yourself!)

Bob

The triangles BED and BFD are congruent as they have the same angles and one side in common.

Therefore BE = BF

On reflection you can probably then proceed by doing a similar job on the intersection of the bisectors of say angle A and angle B.

That way you'll end up with three equal lines which must therefore cross at the same point.  This point is called the in-centre and a cirtcle centred on this point will just touch the three sides tangentially.

There are three ex-centres for any triangle where two external and one internal bisectors meet.

Bob

back later

hi Agnishom

Sorry I had to rush out.  Now for the whole proof.

To Prove:  That the angle bisectors of any triangle meet at a point.  Along the way I shall also show that this point, called the in-centre, is the centre of a circle that just touches the three edges.

Proof.

Let the triangle be ABC.

Let the bisectors of angle B and angle C meet at G.  (see diagram)

Make lines perpendicular to each side from G, meeting the sides at D, E and F as shown.

In triangles BEG and BDG

angle BEG = angle BDG = 90
angle EBG = angle DBG (bisection)
side BG is common.

Therefore, BEG and BDG are congruent.

Therefore GD = GE.

Similarly, triangles CDG and CFG are congruent so

GD = GF

Therefore a circle centre G and radius GD (= GE = GF) will meet the edges of the triangle at D, E  and F at right angles , so the sides are tangent to the circle at these points.

Now draw the line AG and consider the triangles AFG and AEG.

angle AEG = angle AFG = 90
EG = GF (proved above)
AG is common to both.

So AFG and AEG are congruent.

Therefore angle EAG = GAF

Therefore G lies on the bisector of A.

The proof for the circumcircle is almost the same.

When you construct the perpendicular bisector of side AB  to any point G, you make two congruent triangles, AEG and BEG where E is now the midpoint of AB

So GA = GB

So draw a second perpendicular bisector and let it meet the first at G. (rather than G is any point, make it where the lines cross)

Then you can easily show that GA = GB = GC and you're almost finished.

I leave the details as an exercise for you.

Bob

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