Kiepert and Jacobi through Trigonometric Ceva

From Ceva to its trigonometric form

Let $D,E,F$ lie on $BC,CA,AB$ , respectively. We use the trigonometric form of Ceva’s theorem:

AD,BE,CF\text{ are concurrent} \quad\Longleftrightarrow\quad \frac{\sin\angle BAD}{\sin\angle DAC} \frac{\sin\angle CBE}{\sin\angle EBA} \frac{\sin\angle ACF}{\sin\angle FCB} =1.

It follows from ordinary Ceva by applying the sine rule to the three side ratios.

Kiepert’s theorem

Theorem

Construct similarly oriented similar isosceles triangles $XBC,YCA,ZAB$ externally on the sides $BC,CA,AB$ of a triangle $ABC$ . Then the lines $AX,BY,CZ$ are concurrent at a point $K$ , the corresponding Kiepert point.

Triangle ABC with three similarly oriented similar isosceles triangles constructed externally, and the lines AX, BY, CZ meeting at K. — Figure 1. Equal base angles produce Kiepert’s three concurrent lines.

Let $\theta$ be their common base angle. This is the equal-angle case $\alpha=\beta=\gamma=\theta$ of Jacobi’s theorem proved below.

Jacobi’s theorem

The three paired angles need not be equal.

Theorem

Construct the external triangles $XBC,YCA,ZAB$ so that

\angle ZAB=\angle YAC=\alpha, \qquad \angle XBC=\angle ZBA=\beta, \qquad \angle YCA=\angle XCB=\gamma.

Then the lines $AX,BY,CZ$ are concurrent at a point $J$ , called the Jacobi point.

Triangle ABC with three external points X, Y, Z, paired angles alpha, beta, gamma, and the lines AX, BY, CZ meeting at J. — Figure 2. Jacobi’s paired angle conditions give the concurrence of AX, BY, CZ.

Proof

Write $A,B,C$ also for the angles of the triangle. In $BXC$ , the sine rule gives

\frac{BX}{CX}=\frac{\sin\gamma}{\sin\beta}.

Because the construction is external,

\angle ABX=B+\beta, \qquad \angle ACX=C+\gamma.

Applying the sine rule in $ABX$ and $ACX$ therefore gives

\frac{\sin\angle BAX}{\sin\angle XAC} = \frac{\sin\gamma}{\sin\beta} \frac{\sin(B+\beta)}{\sin(C+\gamma)}.

Cyclically,

\frac{\sin\angle CBY}{\sin\angle YBA} = \frac{\sin\alpha}{\sin\gamma} \frac{\sin(C+\gamma)}{\sin(A+\alpha)},

and

\frac{\sin\angle ACZ}{\sin\angle ZCB} = \frac{\sin\beta}{\sin\alpha} \frac{\sin(A+\alpha)}{\sin(B+\beta)}.

Multiplying these three identities gives

\frac{\sin\angle BAX}{\sin\angle XAC} \frac{\sin\angle CBY}{\sin\angle YBA} \frac{\sin\angle ACZ}{\sin\angle ZCB} =1.

Trigonometric Ceva therefore proves that $AX,BY,CZ$ are concurrent at the Jacobi point $J$ .

Setting

\alpha=\beta=\gamma=\theta

gives Kiepert’s theorem.

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Kariya’s theorem

Kariya’s theorem is a direct instance of Jacobi’s paired-angle construction.

Theorem

Let the incircle of $ABC$ , with center $I$ , touch $BC,CA,AB$ at $D,E,F$ , respectively. On the rays $ID,IE,IF$ , beyond $D,E,F$ , choose $X,Y,Z$ so that

IX=IY=IZ.

Then the lines $AX,BY,CZ$ are concurrent.

Triangle ABC with incircle center I and contact points D, E, F; points X, Y, Z lie beyond the contact points with IX, IY, IZ equal, and the dashed lines AX, BY, CZ meet at one point. — Figure 3. Kariya’s construction realizes Jacobi’s paired-angle conditions.

Proof

Reflection in the angle bisector $BI$ maps $D$ to $F$ and the ray $ID$ to the ray $IF$ . Since $IX=IZ$ , it maps $X$ to $Z$ , and hence

\angle XBC=\angle ZBA.

Cyclically,

\angle YCA=\angle XCB, \qquad \angle ZAB=\angle YAC.

These are Jacobi’s three paired-angle conditions, so $AX,BY,CZ$ are concurrent.

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If the common distance is allowed to equal the inradius, then $X=D$ , $Y=E$ , and $Z=F$ ; the concurrence point is the Gergonne point.

From Ceva to its trigonometric form

Kiepert’s theorem

Jacobi’s theorem

Kariya’s theorem

Related notes and diagrams