# Consider acute triangle ABC and points M, N and P the midpoints of minor arcs BC ,CA, respectively AB of Circumscribed circle of triangle ABC .The point I is the circle center inscribed in the...

Consider acute triangle ABC and points M, N and P the midpoints of minor arcs BC ,

CA, respectively AB of Circumscribed circle of triangle ABC .The point I is the circle center inscribed in the triangle ABC.

a) Show that the lines AM, BN and CP are competing in I.

b) Show that the triangle BIM is isosceles .

c) Prove that point I is orthocenter of l triangle MNP.

*print*Print*list*Cite

### 1 Answer

(a) AM is the angle bisector of `angle BAC ` .(M is the bisector of the intercepted arc, so `angle BAM cong angle CAM ` .) Similarly BN and CP are angle bisectors. The point of concurrency of the angle bisectors of a triangle is the incenter (the center of the inscribed circle.)

(b) `m angle IBM ` equals 1/2 the measure of arc MN which is 1/2 the sum of arcs NC and MC.

`m angle BIM ` is a chord-chord angle and is equal to 1/2 the sum of the intercepted arcs or 1/2 (arc AN + arc BM).

But arcs AN and NC are congruent as are arcs MC and BM.

So `m angle IBM = m angle BIM ` and the triangle is isosceles by the converse of the isosceles triangle theorem.

(c) `bar(BN) _|_ bar (PM) `

The measure of angle BJP (where J is the intersection of the chords) is 1/2(arc BP + arc MN)=1/2(arc BP + arc MC + arc CN)

The measure of angle NJP is 1/2(arc BM + arc PN)=1/2(arc BM + arc PA + arc AN)

But the measures of arcs BP=AP, MC=BM, and CN=AN.

Thus `m angle BJP = m angle NJP ==> m angle BJP=90^@ `

A similar argument shows `bar(MA) _|_ bar(PN), bar(PC) _|_ bar(MN) ` .

The orthocenter is the point of concurrency of the altitudes of a triangle, so I is the orthocenter of the triangle.

**Sources:**

### Hide Replies ▲

Thanks a lot!