Trident lemma

The trident lemma , also called the trefoil lemma and the Mansion lemma , is a theorem in the geometry of a triangle related to the properties of inscribed , out-inscribed and circumscribed circles of a triangle.

The trident lemma is used as an auxiliary statement in the proof of many theorems, in particular, the Euler formula or the proof of the existence of the Euler circle .

The name "Mansion's lemma" is in honor of the French mathematician Paul Mansion .

Wording

Trident lemma.

Shamrock Lemma.

Mansion's lemma.

Let in a triangle $A B C$ ${\ displaystyle ABC}$ $ABC$ point $I$ ${\ displaystyle I}$ $I$ - center of the inscribed circle , point $I_{a}$ ${\ displaystyle I_ {a}}$ $I_{a}$ Is the center of the incircle opposite the vertex $A$ ${\ displaystyle A}$ $A$ , and the point $L$ ${\ displaystyle L}$ $L$ - point of intersection of the segment $II_{a}$ ${\ displaystyle II_ {a}}$ $II_{a}$ with an arc of a circumscribed circle (see right). Then the point $L$ ${\ displaystyle L}$ $L$ equidistant from $I$ ${\ displaystyle I}$ $I$ , $I_{a}$ ${\ displaystyle I_ {a}}$ $I_{a}$ , $B$ ${\ displaystyle B}$ $B$ and $C$ ${\ displaystyle C}$ $C$ .

Particular variants of this statement have different names.

Manson's theorem ^[1] : $L$ ${\ displaystyle L}$ $L$ equidistant from $I$ ${\ displaystyle I}$ $I$ and $I_{a}$ ${\ displaystyle I_ {a}}$ $I_{a}$ .
Shamrock Lemma , or the trident lemma ^[2] , or the Mansion lemma ^[3] : $L$ ${\ displaystyle L}$ $L$ equidistant from $I$ ${\ displaystyle I}$ $I$ , $B$ ${\ displaystyle B}$ $B$ and $C$ ${\ displaystyle C}$ $C$ .
Trident lemma ^[4] : $L$ ${\ displaystyle L}$ $L$ equidistant from $I$ ${\ displaystyle I}$ $I$ , $I_{a}$ ${\ displaystyle I_ {a}}$ $I_{a}$ , $B$ ${\ displaystyle B}$ $B$ and $C$ ${\ displaystyle C}$ $C$ .

Another option for setting the point $L$ ${\ displaystyle L}$ $L$ - as the center of the arc $B C$ ${\ displaystyle BC}$ $BC$ circumscribed circle not containing a point $A$ ${\ displaystyle A}$ $A$ ^[3] .

Proof

Under $\angle A,\angle B$ ${\ displaystyle \ angle A, \ angle B}$ $\angle A,\angle B$ we will understand the angles $\angle BAC,\angle ABC$ ${\ displaystyle \ angle BAC, \ angle ABC}$ $\angle BAC,\angle ABC$ respectively. If the beam $A I$ ${\ displaystyle AI}$ $AI$ intersects the circumscribed circle at the point $L$ ${\ displaystyle L}$ then $L$ ${\ displaystyle L}$ is the midpoint of the arc $B C$ ${\ displaystyle BC}$ , section $A L$ ${\ displaystyle AL}$ is the angle bisector $\angle A$ ${\ displaystyle \ angle A}$ . Having drawn a cut $B I$ ${\ displaystyle BI}$ , notice, that

\angle BIL=\angle A/2+\angle B/2,

{\ displaystyle \ angle BIL = \ angle A / 2 + \ angle B / 2,}

because $\angle BIL$ ${\ displaystyle \ angle BIL}$ external to the triangle $\triangle AIB$ ${\ displaystyle \ triangle AIB}$ , and

\angle LBI=\angle LBC+\angle CBI=\angle A/2+\angle B/2,

{\ displaystyle \ angle LBI = \ angle LBC + \ angle CBI = \ angle A / 2 + \ angle B / 2,}

because

\angle LBC

{\ displaystyle \ angle LBC}

and

\angle LAC=\angle A/2

{\ displaystyle \ angle LAC = \ angle A / 2}

are equal, as they rely on one arc

L C

{\ displaystyle LC}

.

Means triangle $\triangle BLI$ ${\ displaystyle \ triangle BLI}$ isosceles, i.e. $BL=LI.$ ${\ displaystyle BL = LI.}$ Equality $CL=BL$ ${\ displaystyle CL = BL}$ follows from the fact that the same angle is based on both of these chords $\angle A/2.$ ${\ displaystyle \ angle A / 2.}$ In this way, $BL=LI=LC.$ ${\ displaystyle BL = LI = LC.}$

We have shown that $BL=LI=LC$ ${\ displaystyle BL = LI = LC}$ . Now we prove that the "handle" of the trident $LI_{a}$ ${\ displaystyle LI_ {a}}$ equal to the same value.

Extend the side $A B$ ${\ displaystyle AB}$ per point $B$ ${\ displaystyle B}$ and take a point somewhere on this continuation $E$ ${\ displaystyle E}$ . Under $\angle A$ ${\ displaystyle \ angle A}$ will understand $\angle BAC,$ ${\ displaystyle \ angle BAC,}$ under $\angle B$ ${\ displaystyle \ angle B}$ we will mean the angle $\angle EBC=180^{\circ }-\angle ABC.$ ${\ displaystyle \ angle EBC = 180 ^ {\ circ} - \ angle ABC.}$

Then we need to understand that the triangle $\triangle BLI_{a}$ ${\ displaystyle \ triangle BLI_ {a}}$ isosceles, that is, that $\angle LBI_{a}=\angle LI_{a}B$ ${\ displaystyle \ angle LBI_ {a} = \ angle LI_ {a} B}$ .

One side,

\angle LBI_{a}=\angle B/2-\angle A/2

{\ displaystyle \ angle LBI_ {a} = \ angle B / 2- \ angle A / 2}

and

\angle EBI_{a}=\angle A/2+\angle BI_{a}A

{\ displaystyle \ angle EBI_ {a} = \ angle A / 2 + \ angle BI_ {a} A}

because

\angle EBI_{a}

{\ displaystyle \ angle EBI_ {a}}

outer in triangle:

\triangle BI_{a}A,

{\ displaystyle \ triangle BI_ {a} A,}

those,

\angle B/2-\angle A/2=\angle BI_{a}A

{\ displaystyle \ angle B / 2- \ angle A / 2 = \ angle BI_ {a} A}

Variations and generalizations

The external trident lemma

Trident lemma for two centers of incircle circles (“external” trident lemma)

Euler Circle Relationship

Through the trident lemma, one can gracefully prove the Euler circle .

Consider an acute-angled triangle ABC. Note that quadrangles $AH_{c}HH_{b}$ ${\ displaystyle AH_ {c} HH_ {b}}$ , $BH_{c}HH_{a}$ ${\ displaystyle BH_ {c} HH_ {a}}$ , $H_{b}HH_{a}C$ ${\ displaystyle H_ {b} HH_ {a} C}$ inscribed (Fig. 1). Therefore, the angles are equal $\measuredangle HH_{b}H_{c}=\measuredangle HAH_{c}=\measuredangle HCH_{a}=\measuredangle HH_{b}H_{a}$ ${\ displaystyle \ measuredangle HH_ {b} H_ {c} = \ measuredangle HAH_ {c} = \ measuredangle HCH_ {a} = \ measuredangle HH_ {b} H_ {a}}$ (pic 2).

picture 1

figure 2

It follows that $H_{b}H$ ${\ displaystyle H_ {b} H}$ - bisector in a triangle $\triangle H_{c}H_{b}H_{a}$ ${\ displaystyle \ triangle H_ {c} H_ {b} H_ {a}}$ . For completely similar reasons $H_{a}H$ ${\ displaystyle H_ {a} H}$ and $H_{c}H$ ${\ displaystyle H_ {c} H}$ also bisectors in this triangle (Fig. 3). You may also notice that $A B,$ ${\ displaystyle AB,}$ $B C,$ ${\ displaystyle BC,}$ $C A$ ${\ displaystyle CA}$ - external bisectors to the triangle $\triangle H_{c}H_{b}H_{a}$ ${\ displaystyle \ triangle H_ {c} H_ {b} H_ {a}}$ (because each of them is perpendicular to its internal bisector). Therefore, we can apply the trident lemma three times, for each side (Fig. 4).

figure 3

figure 4

From this we get that the midpoints of the segments $H A, H B, H C$ ${\ displaystyle HA, HB, HC}$ lie on the circle described near the orthogon . Now we apply the external trident lemma three times (Fig. 5).

figure 5

We get that midpoints $A B, B C, C A$ ${\ displaystyle AB, BC, CA}$ lie on the circle described near the orthogon.

Note

In order to prove the Euler circle for an obtuse triangle $A B C$ ${\ displaystyle ABC}$ with an obtuse angle $A$ ${\ displaystyle A}$ , just consider the acute-angled triangle $B C H$ ${\ displaystyle BCH}$ with orthocenter $A$ ${\ displaystyle A}$ , and apply the same reasoning to it.

Notes

↑ Problem 52395 // "The system of problems in geometry of R. K. Gordin"
↑ Hakobyan A.V. Geometry in pictures .
↑ ¹ ² Emelyanov L.A. Schiffler's point: in memory of I.F. Sharygin . - Mathematics at school, 2006. - No. 6 . - S. 58-60 . - ISSN 0130-9358 .
↑ R.N. Karasev. Problems for the school mathematical circle / R. N. Karasev, V. L. Dolnikov, I. I. Bogdanov, A. V. Akopyan. - S. 4.

[1] Problem 52395 // "The system of problems in geometry of R. K. Gordin"

[2] Hakobyan A.V. Geometry in pictures .

[emelyanov-3] ¹ ² Emelyanov L.A. Schiffler's point: in memory of I.F. Sharygin . - Mathematics at school, 2006. - No. 6 . - S. 58-60 . - ISSN 0130-9358 .

[4] R.N. Karasev. Problems for the school mathematical circle / R. N. Karasev, V. L. Dolnikov, I. I. Bogdanov, A. V. Akopyan. - S. 4.