শেভা'র উপপাদ্য:ত্রিকোণমিতিক রূপ

Ceva's Theorem has multiple forms. One form is the standard one with a product of ratios of lengths equaling one. However, there is another lesser-known form as well: the trigonometric form! It is not hard to derive from the traditional Ceva's Theorem, and in this post we will do so.

Here, we have concurrent cevians $AD$, $BE$, and $CF$ intersecting at a point $P$ inside the triangle. By Ceva's Theorem, we know that $$\dfrac{AF}{FB}\cdot\dfrac{BD}{DC}\cdot\dfrac{CE}{EA}=1.$$ How can we rewrite this in a different form? Think about it for a while before continuing.

We do this with something called the Ratio Lemma, which states that for a cevian $AD$ in $\triangle ABC$, $$\dfrac{BD}{DC}=\dfrac{AB\sin\angle BAD}{AC\sin\angle CAD}.$$ A discussion of greater detail is/will soon be posted here.

In fact, those side lengths will cancel out! Note that from looking at cevians $BE$ and $CF$ we obtain two very similar equations: $$\begin{align*}\dfrac{CE}{EA}&=\dfrac{BC\sin\angle CBE}{BA\sin\angle ABE},\\\dfrac{AF}{FB}&=\dfrac{CA\sin\angle ACF}{CB\sin\angle BCF}.\end{align*}$$ Therefore by plugging into Ceva's Theorem we obtain $$\begin{align*}\dfrac{AF}{FB}\cdot\dfrac{BD}{DC}\cdot\dfrac{CE}{EA}&=\dfrac{AB\sin\angle BAD}{AC\sin\angle CAD}\cdot \dfrac{BC\sin\angle CBE}{BA\sin\angle ABE}\cdot\dfrac{CA\sin\angle ACF}{CB\sin\angle BCF}\\&= \dfrac{\sin\angle BAD}{\sin\angle CAD}\cdot \dfrac{\sin\angle CBE}{\sin\angle ABE}\cdot\dfrac{\sin\angle ACF}{\sin\angle BCF}\\\implies 1&=\dfrac{\sin\angle BAD}{\sin\angle CAD}\cdot \dfrac{\sin\angle CBE}{\sin\angle ABE}\cdot\dfrac{\sin\angle ACF}{\sin\angle BCF}.\end{align*}$$

This last equality is the trigonometric form of Ceva's Theorem!

Confused? Read over the proof a few times until you understand it. Then I'll give a few problems that can be solved using the trig form.

Here are some problems that utilize the formula we just discovered. With the exception of the first one, these are all fairly difficult. Do not get discouraged if you don't get far in them!

PROBLEM ONE: In $\triangle ABC$, points $D$, $E$, and $F$ are situated on sides $AB$, $BC$, $CA$ respectively such that $\sin\angle BAD=2\sin\angle CAD$ and $\sin\angle ABE=3\sin\angle CBE$. If $AD$, $BE$, and $CF$ intersect at a point $P$ inside $\triangle ABC$, find the ratio $\sin\angle ACF/\sin\angle BCF$.

PROBLEM TWO: In $\triangle ABC$, let $D$ and $E$ reside on $AB$ and $AC$ respectively. Suppose that $\angle BAD=15^\circ$, $\angle CAD=35^\circ$, $\angle ABE=20^\circ$, and $\angle CBE=40^\circ$. Let $P$ be the intersection point of segments $AD$ abd $BE$. Given that $P$ lies inside $\triangle ABC$, find $|\angle ACP-\angle BCP|$. (You can, and probably should, use a calculator).

PROBLEM THREE (USAMO 1997): Let $ABC$ be a triangle. Take points $D$, $E$, $F$ on the perpendicular bisectors of $BC$, $CA$, $AB$ respectively. Show that the lines through $A$, $B$, $C$ perpendicular to $EF$, $FD$, $DE$ respectively are concurrent.

PROBLEM FOUR (Unknown source): Points $A_1,B_1,C_1$ are chosen on the sides $BC,CA,AB$ respectively of a triangle $ABC$. Denote by $G_a,G_b,G_c$ the centroids of triangles $AB_1C_1$, $BC_1A_1$, $CA_1B_1$ respectively. Prove that the lines $AG_a$, $BG_b$, $CG_c$ are concurrent if and only if lines $AA_1,BB_1,CC_1$ are concurrent.

I recently came across with this post an hour ago. Here I would like to share my own method in proving the Trigonometric Ceva theorem. Let me start with a lemma.

Lemma: Let $ABC$ be a triangle, and let $P$ be a varaible inside $ABC$. Let $B_1$, $B_2$ be varaible point on segment $AC$ let circumcircles $AB_1P$ and $AB_2P$ intersect $AB$ at $C_1$ and $C_2$ respectively. Then $\displaystyle \frac{B_1B_2}{C_1C_2}=\frac{\sin \angle PAC}{\sin \angle PAB}$.

Proof of Lemma: The obvious move is by using sine rule. Without loss of generality, $B_1$ lies betwwen $A$ and $B_2$, then we have $C_2$ lies between $A$ and $C_1$. We have $\displaystyle \frac{\sin \angle PAC}{\sin \angle PAB}=\frac{B_2P\sin \angle PB_2A}{AP}\cdot \frac{AP}{C_2P\sin \angle PC_2A}=\frac{B_2P\sin \angle PB_2A}{C_2P\sin \angle PC_2A}=\frac{B_2P\sin \angle PC_2C_1}{C_2P\sin \angle PB_1B_2}=\frac{B_1B_2\sin \angle C_2PC_1}{C_1C_2\sin \angle B_2PB_1}$. Let us prove that in fact $\angle B_2PB_1=\angle C_2PC_1$.

$\angle B_2PB_1=\angle APB_2-\angle APB_1=\angle AC_2B_2-\angle AC_1B_1=180-\angle TC_2C_1-\angle C_2C_1T=\angle C_2TC_1=\angle B_1TB_2$. We get $B_2, B_1, T, P$ concyclic. Similarly, $C_2, C_1, T, P$ concyclic. So $\angle B_2PB_1=\angle B_1TB_2=\angle C_2TC_1=\angle C_2PC_1$. The lemma is proved.

Let us proceed to the main theorem.

Proof of the theorem: Construct similar points $A_1$ and $A_2$ on $BC$ so that $B_1PA_1C$ and $B_2PA_2C$ are cyclic. Then by Miquel's theorem, $CC_1PA_1$ and $CC_2PA_2$ are also cyclic. So, by the lemma, we get $$\frac{\sin \angle PAC}{\sin \angle PAB}\cdot \frac{\sin \angle PBA}{\sin \angle PBC}\cdot \frac{\sin \angle PCB}{\sin \angle PCA}=\frac{B_1B_2}{C_1C_2} \cdot \frac{C_1C_2}{A_1A_2} \cdot \frac{A_1A_2}{B_1B_2}=1$$

as required. Q.E.D.

Hope you like the proof :)

Problem. (ISL 2001 G1.) Let $A_1$ be the center of the square inscribed in acute triangle $ABC$ with two vertices of the square on side $BC$. Thus one of the two remaining vertices of the square is on side $AB$ and the other is on $AC$. Points $B_1,\ C_1$ are defined in a similar way for inscribed squares with two vertices on sides $AC$ and $AB$, respectively. Prove that lines $AA_1,\ BB_1,\ CC_1$ are concurrent.

Let $~$ $\square{KLMN}$ $~$ be the square with center $~$ $A_1$ $~$ such that $~$ $K \in \overline{AB}\; \wedge\; N \in \overline{AC}$ $~$.

$~$ $A_1$ $~$ is the center of $~$ $\square{KLMN}\; \Longrightarrow\; \boxed{ \overline{A_1K} = \overline{A_1N} }\; \wedge\; \boxed{ \angle{A_1KN} = 45^\circ = \angle{A_1NK} }\quad \cdots\, (1)$

$\square{KLMN}$ $~$ is a square $~$ $\Longrightarrow\; KN \parallel LM\; \Longrightarrow\; KN \parallel BC\; \Downarrow$

$\Longrightarrow\; \boxed{ \angle{AKN} = \angle{ABC} = \angle{B} }\; \wedge\; \boxed{ \angle{ANK} = \angle{ACB} = \angle{C} }\quad \cdots\, (2)$

In $~$ $\triangle{AKA_1}\; \wedge\; \triangle{ANA_1}\; \stackrel{ \text{Law of sines} }{\Downarrow}$

$\Longrightarrow\; \frac { \sin \angle{KAA_1} } { \sin \angle{A_1KA} } = \frac { \overline{A_1K} } { \overline{A_1A} } \stackrel { (1) } {=} \frac { \overline{A_1N} } { \overline{A_1A} } = \frac { \sin \angle{NAA_1} } { \sin \angle{A_1NA} }\; \Downarrow$

$\Longrightarrow\; \frac { \sin \angle{KAA_1} } { \sin \angle{NAA_1} } = \frac { \sin \angle{A_1KA} } { \sin \angle{A_1NA} } \stackrel { (1)\; \wedge\; (2) } {=} \frac { \sin \angle{ B + 45^\circ } } { \sin \angle{ C + 45^\circ } }\; \Downarrow$

$\Longrightarrow\; \boxed{ \frac { \sin \angle{BAA_1} } { \sin \angle{CAA_1} } = \frac { \sin \angle{ B + 45^\circ } } { \sin \angle{ C + 45^\circ } } }\quad \cdots\, (3)$

Similarly $~$ $\boxed{ \frac { \sin \angle{CBB_1} } { \sin \angle{ABB_1} } = \frac { \sin \angle{ C + 45^\circ } } { \sin \angle{ A + 45^\circ } } }\; \wedge\; \boxed{ \frac { \sin \angle{ACC_1} } { \sin \angle{BCC_1} } = \frac { \sin \angle{ A + 45^\circ } } { \sin \angle{ B + 45^\circ } } }\quad \cdots\, (4)$

$\stackrel { (3)\; \wedge\; (4) } { \Longrightarrow }\; \boxed { \frac { \sin \angle{BAA_1} } { \sin \angle{CAA_1} } \cdot \frac { \sin \angle{ACC_1} } { \sin \angle{BCC_1} } \cdot \frac { \sin \angle{CBB_1} } { \sin \angle{ABB_1} } = }$

$= \frac { \sin \angle{ B + 45^\circ } } { \sin \angle{ C + 45^\circ } } \cdot \frac { \sin \angle{ A + 45^\circ } } { \sin \angle{ B + 45^\circ } } \cdot \frac { \sin \angle{ C + 45^\circ } } { \sin \angle{ A + 45^\circ } } =$

$\boxed{ = 1 } \; \stackrel { \text{trigonometric form of Ceva's Theorem} } { \Downarrow}$

Lines $~$ $AA_1,\, BB_1\; \wedge\; CC_1$ are concurrent. $~$ $\blacksquare$

$-$ Solution by AoPS user Max D.R.