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MIDLINES OF QUADRILATERAL
Резюме. In this section we deal with an interesting fact which will imply \(A B C D\) some well known theorems. Namely we prove that in a convex quadrilateral whose diagonals meet at O, the midline of the triangle \(\triangle B O C\) parallel to \(B C\), the midline of the triangle \(\triangle A O D\) parallel to \(A D\), and the midline of the quadrilateral \(A B C D\) connecting the sides \(A B\) and \(C D\) meet at one point, or are parallel.
Ключови думи: Ceva; Menelaus; Stewartes; cevian; concurrency; collinearity; Fermat; Torricelli
1. INTRODUCTION
Observing the history of geometry, we can see that one of the most attractive problems was the concurrency of three lines, and collinearity of three points. Two astonishing results connecting those two problems are Ceva and Menelaus theorems, well known to most of the mathematicians, especially competitive ones. Here we present a theorem in quadrilateral, implying both of those theorems and also Van Aubels theorem which is well conected to those two. We also present interesting collinearity and concurrency in quadrilateral, and another equaivalent form of Ceva’s theorem which sometimes yields the desired result faster than its standard form. The main point is that these results are derived using only similarity and midlines of a triangles which are well known to the primary school students as well as the beginners of high school students. Concurency of three midlines in quadrilateral, as we will see, yields various results and identities. Once the main theorem is proved we can use the identities obtained by itself to prove Stewart’s theorem which yields the lenghts of various cevians, such as a median, an altitude, an angle bisector etc. Depending on what kind of quadrilateral we choose, we can get any point inside the triangle, even extremal points, such as Fermat-Torricelli, centroid and similar ones.
Assumption. Let \(A B C D\) be a convex quadrilateral whose diagonals \(A C\) and \(B D\) meet at \(O\). Let \(P, Q, R, S, M, N\) be the midpoints of the segments \(A O, B O, C O, D O, A B, C D\) respectively. Let \(M N\), meet \(Q R, A C, B D\) at the points \(V, T, U\) respectively. Let also \(V O\) meet \(P Q, Q M\) and \(A B\) at the points \(K, L\) and \(Z\) respectively.
2. LIST OF IDENTITIES 1
(1) \(N R=\tfrac{D O}{2}\)
Proof:
Since \(N\) is the midpoint of \(C D\), and \(R\) is the midpoint of \(C O\), thus \(N R\) is a midline of triangle \(\triangle C O D\), hence \(N R=\tfrac{D O}{2}\)
(2) \(S N=\tfrac{C O}{2}\)
Proof:
Since \(N\) is the midpoint of \(C D\), and \(S\) is the midpoint of \(D O\), thus \(N S\) is a midline of triangle \(\triangle C O D\), hence \(S N=\tfrac{C O}{2}\)
\((3) \text{ } M Q=\tfrac{A O}{2}\)
Proof:
Since \(M\) is the midpoint of \(A B\), and \(Q\) is the midpoint of \(B O\), thus \(M Q\) is a midline of triangle \(\triangle B O A\), hence \(M Q=\tfrac{A O}{2}\) (4) \(S Q=\tfrac{B D}{2}\)
Proof:
Since \(S\) is the midpoint of \(D O\), and \(Q\) is the midpoint of \(B O\), thus
\(S Q=S O+Q O=\tfrac{D O}{2}+\tfrac{B O}{2}=\tfrac{D O+B O}{2}=\tfrac{B D}{2}\)
(5) \(U Q=\tfrac{B D \cdot A O}{2 A C}\)
Proof:
Since \(S N\) is the midline of \(\triangle C O D\) we have \(S N\|C O \Rightarrow S N\| A C\).
Since \(M Q\) is the midline of \(\triangle B O A\) we have \(M Q\|A O \Rightarrow M Q\| A C\).
Hence we have \(S N \| M Q\), so \(\triangle S N U \sim \triangle M U Q\), so we have
\[ \begin{aligned} & \tfrac{U S}{U Q}=\tfrac{N S}{M Q} \Rightarrow \tfrac{U S}{U Q}+1=\tfrac{N S}{M Q}+1 \Rightarrow \tfrac{S Q}{U Q}=\tfrac{N S+M Q}{M Q} \Rightarrow \\ & U Q=\tfrac{S Q \cdot M Q}{S N+M Q}={ }^{(4),(3)} \tfrac{\tfrac{B D}{2} \cdot \tfrac{A O}{2}}{\tfrac{C O}{2}+\tfrac{A O}{2}}=\tfrac{B D \cdot A O}{2 A C} \end{aligned} \]
(6) \(U O=\tfrac{A O \cdot D O-C O \cdot B O}{2 A C}\)
Proof: If UQ<QO then this case is impossible. So let it be UQ>QO \(U O=U Q-Q O={ }^{(5)} \tfrac{B D \cdot A O}{2 A C}-\tfrac{B O}{2}=\tfrac{B D \cdot A O-A C \cdot B O}{2 A C}=\)
\(=\tfrac{(B O+D O) A O-(A O+C O) B O}{2 A C}=\tfrac{A O \cdot D O-B O \cdot C O}{2 A C}\)
(7) \(U S=\tfrac{B D \cdot C O}{2 A C}\) \(U S=Q S-U Q={ }^{(4),(5)} \tfrac{B D}{2}-\tfrac{B D \cdot A O}{2 A C}=\tfrac{B D}{2 A C}(A C-A O)=\tfrac{B D \cdot C O}{2 A C}\)
Proof:
(8) \(P R=\tfrac{A C}{2}\)
Proof:
Since \(P\) is the midpoint of \(A O\), and \(R\) is the midpoint of \(C O\), thus \(P R=P O+R O=\tfrac{A O}{2}+\tfrac{C O}{2}=\tfrac{A O+C O}{2}=\tfrac{A C}{2}\)
(9) \(T R=\tfrac{A C \cdot D O}{2 B D}\)
Proof:
Since \(R N\) is the midline of \(\triangle C O D\) we have \(R N\|D O \Rightarrow R N\| B D\) Since \(M P\) is the midline of \(\triangle B O A\) we have \(M P\|B O \Rightarrow M P\| B D\) Hence we have \(R N \| M P\), so \(\triangle R T N \sim \triangle P M T\), so we have \(\tfrac{P T}{R T}=\tfrac{P M}{N R} \Rightarrow \tfrac{P T}{R T}+1=\tfrac{P M}{N R}+1 \Rightarrow \tfrac{P R}{R T}=\tfrac{P M+N R}{N R} \Rightarrow\)
\(R T=\tfrac{P R \cdot N R}{P M+N R}={ }^{(8),(1)} \tfrac{\tfrac{A C}{2} \cdot \tfrac{D O}{2}}{\tfrac{B O}{2}+\tfrac{D O}{2}}=\tfrac{A C \cdot D O}{2 B D}\)
(10) \(P T=\tfrac{A C \cdot B O}{2 B D}\)
Proof: \( P T=P R-R T={ }^{(9)} \tfrac{A C}{2}-\tfrac{A C \cdot D O}{2 B D} \Rightarrow \tfrac{A C}{2}(B D-D O) \Rightarrow \) \(PT=\tfrac{AC\cdot BO}{2BD}\)
\(\text { (11) } T O=\tfrac{A O \cdot D O-B O \cdot C O}{2 B D} \)
\(T O=P O-P T={ }^{(10)} \tfrac{A O}{2}-\tfrac{A C \cdot B O}{2 B D}=\tfrac{B D \cdot A O-A C \cdot B O}{2 B D}= \) \(=\tfrac{(B O+D O) \cdot A O-(A O+C O) \cdot B O}{2 B D}=\tfrac{A O \cdot D O-B O \cdot C O}{2 B D}\)
\((12)\text{ }PM=\cfrac{BO}{2}\)
Proof:
Since \(M\) is the midpoint of \(A B\), and \(P\) is the midpoint of \(A O\), thus \(A O\) is a midline of triangle \(\triangle B O A\), hence \(P M=\tfrac{B O}{2} P M=\tfrac{B O}{2}\)
(13) \(P S=\tfrac{A D}{2}\)
Proof:
Since \(P\) is the midpoint of \(A O\) and \(S\) is the midpoint of \(D O\), then \(P S\) is a midline of triangle \(\triangle D O A\), so
\(P S=\tfrac{A D}{2}\)
(14) \(N U=\tfrac{M N \cdot C O}{A C}\)
Proof:
Since \(S N \| M Q\), we have \(\triangle S U N \sim \triangle M U Q\), so \[ \begin{aligned} & \tfrac{M U}{N U}=\tfrac{M Q}{S N} \Rightarrow \tfrac{M U}{N U}+1=\tfrac{M Q}{S N}+1 \Rightarrow \tfrac{M N}{U N}=\tfrac{M Q+S N}{S N} \Rightarrow \\ & U N=\tfrac{M N \cdot S N}{M Q+S N}=(2),(3) \tfrac{M N \cdot \tfrac{C O}{2}}{\tfrac{A O}{2}+\tfrac{C O}{2}}=\tfrac{M N \cdot C O}{A C} \end{aligned} \]
(15) \(M T=\tfrac{M N \cdot B O}{B D}\)
Proof:
Since \(P M \| R N\), we have \(\triangle P T M \sim \triangle R T N\)
\(\tfrac{N T}{M T}=\tfrac{N R}{P M} \Rightarrow \tfrac{N T}{M T}+1=\tfrac{N R}{P M}+1 \Rightarrow \tfrac{M N}{M T}=\tfrac{N R+P M}{P M} \Rightarrow\)
\(M T=\tfrac{M N \cdot P M}{N R+P M}=\) (1),(12) \(^{\tfrac{M N}{D O}} \tfrac{\tfrac{B O}{2}}{\tfrac{B O}{2}}=\tfrac{M N \cdot B O}{B D}\)
(16) \(R Q=\tfrac{B C}{2}\)
Proof:
Since \(R\) is the midpoint of \(C O\), and \(Q\) is the midpoint of \(B O\), thus \(R Q\) is a midline of triangle \(\triangle B O C\), hence \(R Q=\tfrac{B C}{2}\)
(17) \(P Q=\tfrac{A B}{2}\)
Proof:
Since \(P\) is the midpoint of \(A O\), and \(O\) is the midpoint of \(B O\), thus \(P Q\) is a midline of a triangle \(\triangle B O A\), hence \(P Q=\tfrac{A B}{2}\)
3. MAIN THEOREM
Theorem 1. Let \(A B C D\) be a convex quadrilateral whose diagonals \(A C\) and \(B D\) meet at \(O\). Let \(P, Q, R, S, M, N\) be the midpoints of the segments \(A O, B O, C O, D O, A B, C D\) respectively. Then the lines \(P S, Q R\) and \(M N\) meet at one point, or are parallel.
Proof:
Let \(A D \| B C\) and let the line parallel to \(B C\) meets \(A C\) and \(A B\) at the points \(H\) and \(M_{1}\) respectively. Since \(N\) is the midpoint of \(C D\) and \(N H \| A D\), then \(H N\) is a midline of triangle \(\triangle A C D\), hence the point \(H\) is a midpoint of \(A C\). Since \(H\) is the midpoint of \(A C\) and \(H M_{1} \| B C\) then \(H M_{1}\) is a midline of triangle \(\triangle A B C\), hence the point \(M_{1}\) is the midpoint of \(A B\). So we have \(M_{1} \equiv M\), so the lines \(A D, B C, M N\) are parallel. Since \(P S\) is the midline of triangle \(\triangle A D O\), then \(P S \| A D\). Since \(Q R\) is a midline of triangle \(\triangle B O C\), then \(Q R \| B C\). So we get the lines \(P S, Q R\) and \(M N\) are parallel. Let \(A D\) be non-parallel to \(B C\).
Pic.1
Pic. 2
We will consider the case when \(A_{\triangle A O B} \gt A_{\triangle C O D}\). The other case is similar. Let \(M N\) meet \(Q R, A C, B D\) at the points \(V, T, U\) respectively.
\[ \begin{aligned} \tfrac{V M}{V N} & =\tfrac{\tfrac{M N \cdot A O \cdot B O}{A O \cdot B O-C O \cdot D O}}{\tfrac{M N \cdot C O \cdot D O}{A O \cdot B O-C O \cdot D O}}=\tfrac{A O \cdot B O}{C O \cdot D O} \Rightarrow \tfrac{V M}{V N}-1=\tfrac{A O \cdot B O}{C O \cdot D O}-1 \Rightarrow \\ \tfrac{M N}{V N} & =\tfrac{A O \cdot B O-C O \cdot D O}{C O \cdot D O} \Rightarrow \\ V N & =\tfrac{M N \cdot C O \cdot D O}{A O \cdot B O-C O \cdot D O} \end{aligned} \] Suppose on the other side that \(P S\) and \(M N\) meet at \(W\). Then similarly \[ W N=\tfrac{M N \cdot C O \cdot D O}{A O \cdot B O-C O \cdot D O} \Rightarrow V \equiv W \] In the former calculations we will denote this point by \(V\).
4. LIST OF IDENTITIES 2
(18) \(V S=\tfrac{A D \cdot B D \cdot C O}{2(A O \cdot B O-C O \cdot D O)}\)
Proof:
Since \(P M \| S U\), we have \(\triangle V U S \sim \triangle V M P\), so \(\tfrac{V P}{V S}=\tfrac{P M}{U S} \Rightarrow \tfrac{V P}{V S}-1=\tfrac{P M}{U S}-1 \Rightarrow \tfrac{P S}{V S}=\tfrac{P M-U S}{U S} \Rightarrow\)
\(V S=\tfrac{P S \cdot U S}{P M-U S}=(7),(12),(13) \tfrac{\tfrac{A D}{2} \cdot \tfrac{B D \cdot C O}{2 A C}}{\tfrac{B O}{2}-\tfrac{B D \cdot C O}{2 A C}}=\tfrac{A D \cdot B D \cdot C O}{2(A C \cdot B O-B D \cdot C O)}=\) \(=\tfrac{A D \cdot B D \cdot C O}{2(A O+C O) B O-2(B O+D O) C O}=\tfrac{A D \cdot B D \cdot C O}{2(A O \cdot B O-C O \cdot D O}\)
(19) \(V R=\tfrac{A C \cdot B C \cdot D O}{2(A O \cdot B O-C O \cdot D O)}\)
Proof:
Since \(Q M \| R T\), we have \(\Delta V R T \sim \Delta V M Q\), so
\(\tfrac{V Q}{V R}=\tfrac{Q M}{R T} \Rightarrow \tfrac{V Q}{V R}-1=\tfrac{Q M}{R T}-1 \Rightarrow \tfrac{R Q}{V R}=\tfrac{Q M-R T}{R T} \Rightarrow\)
\(V R=\tfrac{R T \cdot R Q}{Q M-R T}={ }^{(9),(3),(16)} \tfrac{\tfrac{A C \cdot D O}{2 B D} \cdot \tfrac{B C}{2}}{\tfrac{A O}{2}-\tfrac{A C \cdot D O}{2 B D}}=\tfrac{A C \cdot B C \cdot D O}{2(B D \cdot A O-A C \cdot D O)}=\) \(=\tfrac{A C \cdot B C \cdot D O}{2(B O+D O) A O-(A O+C O) D O}=\tfrac{A C \cdot B C \cdot D O}{2(A O \cdot B O-C O \cdot D O)}\)
(20) \(V Q=\tfrac{B D \cdot B C \cdot A O}{2(A O \cdot B O-C O \cdot D O)}\)
Proof:
\(V Q=V R+R Q={ }^{(19),(16)} \tfrac{A C \cdot B C \cdot D O}{2(A O \cdot B O-C O \cdot D O)}+\tfrac{B C}{2}=\)
\(=\tfrac{B C}{2(A O \cdot B O-C O \cdot D O)}(A C \cdot D O+A O \cdot B O-C O \cdot D O)=\) \[ \begin{aligned} & \tfrac{B C}{2(A O \cdot B O-C O \cdot D O)}(A O \cdot D O+C O \cdot D O+A O \cdot B O-C O \cdot D O)= \\ & =\tfrac{B C}{2(A O \cdot B O-C O \cdot D O)}(A O \cdot D O+A O \cdot B O)=\tfrac{B C \cdot B D \cdot A O}{2(A O \cdot B O-C O \cdot D O)} \\ & (21) \quad V P=\tfrac{A D \cdot A C \cdot B O}{2(A O \cdot B O-C O \cdot D O)} \end{aligned} \]
Proof:
\[ \begin{aligned} & V P=P S+S V=(13),(18) \tfrac{A D}{2}+\tfrac{A D \cdot B D \cdot C O}{2(A O \cdot B O-C O \cdot D O)}= \\ & =\tfrac{A D}{2(A O \cdot B O-C O \cdot D O)}(A O \cdot B O-C O \cdot D O+B O \cdot C O+C O \cdot D O) \\ & \quad=\tfrac{A C \cdot A D \cdot B O}{2(A O \cdot B O-C O \cdot D O)} \end{aligned} \]
(22) \(Q L=\tfrac{B D \cdot A O \cdot C O}{2 A C \cdot D O}\)
Proof:
Since \(R O \| Q L\) we have \(\triangle V R O \sim \triangle V L Q\), so we have \(\tfrac{Q L}{R O}=\tfrac{V Q}{V R} \Rightarrow Q L=(19),(20) \tfrac{R O \cdot V Q}{V R}=\tfrac{\tfrac{C O}{2} \cdot \tfrac{B D \cdot A O}{2(A O \cdot B O-C O \cdot D O)}}{\tfrac{A C \cdot B C \cdot D O}{2(A O \cdot B O-C O \cdot D O)}}=\) \(=\tfrac{B D \cdot B C \cdot A O \cdot C O}{2 A C \cdot B C \cdot D O}=\tfrac{B D \cdot A O \cdot C O}{2 A C \cdot D O}\)
(23) \(K P=\tfrac{A C \cdot A B \cdot D O}{2(B D \cdot C O+A C \cdot D O)}\)
Proof:
Since \(P O \| Q L\) we have \(\triangle P O K \sim \triangle L K Q\), so we have \(\tfrac{Q K}{K P}=\tfrac{Q L}{P O} \Rightarrow \tfrac{Q K}{K P}+1=\tfrac{Q L}{P O}+1 \Rightarrow \tfrac{P Q}{K P}=\tfrac{Q L+P O}{P O} \Rightarrow\)
\(K P=\tfrac{P O \cdot P Q}{Q L+P O}=(22),(17) \tfrac{\tfrac{A O}{2} \cdot \tfrac{A B}{2}}{\tfrac{B D \cdot A O \cdot C O}{2 A C \cdot D O}+\tfrac{A O}{2}}=\tfrac{A C \cdot A B \cdot D O}{2(B D \cdot C O+A C \cdot D O)}\)
(24) \(Q K=\tfrac{B D \cdot A B \cdot C O}{2(B D \cdot C O+A C \cdot D O)}\)
Proof:
Since \(P O \| Q L\) we have \(\triangle P O K \sim \triangle L K Q\), so we have
\(Q K=P Q-K P={ }^{(17),(23)} \tfrac{A B}{2}-\tfrac{A C \cdot A B \cdot D O}{2(B D \cdot C O+A C \cdot D O)}=\)
\(=\tfrac{A B(B D \cdot C O+A C \cdot D O-A C \cdot D O)}{2(B D \cdot C O+A C \cdot D O)}=\tfrac{B D \cdot A B \cdot C O}{2(B D \cdot C O+A C \cdot D O)}\)
(25) \(V N=\tfrac{M N \cdot C O \cdot D O}{A O \cdot B O-C O \cdot D O}\)
Proof:
Since \(N R\) is the midline of triangle \(\triangle C O D\), thus \(N R\left\|B D \Rightarrow{ }^{(1)} N R\right\| U Q\). So we have similarity \(\triangle V U Q \sim \triangle V N R\), thus
\(\tfrac{V U}{V N}=\tfrac{U Q}{N R} \Rightarrow \tfrac{V U}{V N}-1=\tfrac{U Q}{N R}-1 \Rightarrow \tfrac{U N}{V N}=\tfrac{U Q-N R}{N R}={ }^{(1),(5)}\)
\(=\tfrac{\tfrac{B D \cdot A O}{2 A C}-\tfrac{D O}{2}}{\tfrac{D O}{2}}=\tfrac{B D \cdot A O-A C \cdot D O}{A C \cdot D O}=\tfrac{(B O+D O) A O-(A O+C O) D O}{A C \cdot D O}=\)
\(=\tfrac{A O \cdot B O-C O \cdot D O}{A C \cdot D O} \Rightarrow\)
\(V N=\tfrac{U N \cdot A C \cdot D O}{A O \cdot B O-C O \cdot D O}=\tfrac{\tfrac{M N \cdot C O}{A C} \cdot A C \cdot D O}{A O \cdot B O-C O \cdot D O}=\tfrac{M N \cdot C O \cdot D O}{A O \cdot B O-C O \cdot D O}\)
(26) \(V M=\tfrac{M N \cdot A O \cdot B O}{A O \cdot B O-C O \cdot D O}\)
Proof:
\[ V M=V N+M N={ }^{(25)} \tfrac{M N \cdot C O \cdot D O}{A O \cdot B O-C O \cdot D O}+M N=\tfrac{M N \cdot A O \cdot B O}{A O \cdot B O-C O \cdot D O} \] 5. SUBMAIN THEOREM
Theorem 2.Let \(A B C D\) be a convex quadrilateral whose diagonals \(A C\) and \(B D\) meet at \(O\). Let \(P, Q, R, S, M, N\) be the midpoints of the segments \(A O, B O, C O, D O, A B, C D\) respectively. Let \(A D\) not parallel to \(B C\) and let the lines \(P S, Q R\) and \(M N\) meet at a point \(V\). Let the lines \(A D\) and \(B C\) meeet at \(G\). Then the points \(G, V, O\) are collinear.
Pic. 3
Proof:
Let the line \(O V\) meet \(B C\) at \(G_{1}\). Since \(Q R\) is a midline of triangle \(\triangle B O C\), then \(Q R \| B C\) so we have \(Q V \| B G_{1}\). Since \(Q\) is the midpoint of \(O B\) and \(Q V \| B G_{1}\)
then \(Q V\) is a midline of triangle \(\triangle B O G_{1}\), hence \(V\) is midpoint of \(O G_{1}\).
If we suppose the line \(O V\) meets \(A D\) at the point \(G_{2}\), similarly we obtain that the point \(V\) is midpoint of \(O G_{2}\).
So we have \(G_{1} \equiv G_{2} \equiv G\).
After we have proved Theorem 2, we can see that the triangles \(\triangle A B G\) and \(\triangle P V Q\) are homothetic with respect to \(O\) and the absolute value of the coefficient of homotethy is 2 .
We will now show that the consequences of these theorems are some well known theorems. The converses of those theorems (if exist) are trivial, so we will not deal with them
6. CONSEQUENCES
Corolary 1 (Menelaus theorem)
Let the line p meets segments \(A B, B C\) and the extension of the segment \(A C\) at the points \(C_{1}, A_{1}\) and \(B_{1}\) respectively.Then
\(\tfrac{A B_{1}}{B_{1} C} \cdot \tfrac{C A_{1}}{A_{1} B} \cdot \tfrac{B C_{1}}{C_{1} A}=1\)
Proof:
Consider the line \(O G\) and the triangle \(\triangle A D B\). It suffices to prove
\(\tfrac{A Z}{Z B} \cdot \tfrac{B O}{O D} \cdot \tfrac{D G}{A G}=1\)
Using identities (23) and (24); (18) and (21) we have
\(\tfrac{2 P K}{2 K Q} \cdot \tfrac{B O}{O D} \cdot \tfrac{2 \mathrm{VS}}{2 P V}=\) \(\cfrac{\tfrac{A C \cdot A B \cdot D O}{2(B D \cdot C O+A C \cdot D O)}}{\tfrac{B D \cdot A B \cdot C O}{2(B D \cdot C O+A C \cdot D O)}} \cdot \tfrac{B O}{O D} \cdot \cfrac{\tfrac{A D \cdot B D \cdot C O}{2(A O \cdot B O-C O \cdot D O)}}{\tfrac{A D \cdot A C \cdot B O}{2(A O \cdot B O-C O \cdot D O)}}\) \(=\tfrac{A C \cdot D O}{B D \cdot C O} \cdot \tfrac{B O}{O D} \cdot \tfrac{B D \cdot C O}{A C \cdot B O}=1\),
which completes the proof.
Corolary 2 (Cevas theorem)
Let \(A_{1}, B_{1}, C_{1}\) be the points on the segments \(B C, C A, A B\) respectively such that \(A A_{1}, B B_{1}\) and \(C C_{1}\) meet at one point. Then
\(\tfrac{A B_{1}}{B_{1} C} \cdot \tfrac{C A_{1}}{A_{1} B} \cdot \tfrac{B C_{1}}{C_{1} A}=1\)
Proof:
Consider the point \(O\) and the triangle \(\triangle G A B\). It suffices to prove \(\tfrac{A Z}{Z B} \cdot \tfrac{B C}{C G} \cdot \tfrac{G D}{D A}=1\).
Using identities (23) and (24); (19), (18) we have \(\tfrac{2 P K}{2 K Q} \cdot \tfrac{B C}{2 V R} \cdot \tfrac{2 V S}{D A}=\)
\(=\tfrac{\tfrac{A C \cdot A B \cdot D O}{2(B D \cdot C O+A C \cdot D O)}}{\tfrac{B D \cdot A B \cdot C O}{2(B D \cdot C O+A C \cdot D O)}} \cdot \tfrac{B C}{\tfrac{A C \cdot B C \cdot D O}{A O \cdot B O-C O \cdot D O}} \cdot \tfrac{\tfrac{A D \cdot B D \cdot C O}{A O \cdot B O-C O \cdot D O}}{D A}=1\),
which completes the proof.
Corolary 3 (Van Aubels theorem)
Let \(A_{1}, B_{1}, C_{1}\) be the points on the segments \(B C, C A, A B\) respectively such that \(A A_{1}, B B_{1}\) and \(C C_{1}\) meet at one point \(O\).Then
\(\tfrac{A O}{O A_{1}}=\tfrac{A B_{1}}{B_{1} C}+\tfrac{A C_{1}}{C_{1} B}\)
Proof:
Consider the point \(O\) and the triangle \(\triangle A B G\). It suffices to prove \(\tfrac{A O}{O C}=\tfrac{A Z}{Z B}+\tfrac{A D}{D G}\)
Using identities(23), (24), (18) we have
\(\tfrac{A Z}{Z B}+\tfrac{A D}{D G}=\tfrac{2 P K}{2 K Q}+\tfrac{A D}{2 V S}=\tfrac{\tfrac{A C \cdot A B \cdot D O}{B D \cdot C O+A C \cdot D O}}{\tfrac{B D \cdot A B \cdot C O}{B D \cdot C O+A C \cdot C O}}+\tfrac{A D}{\tfrac{A D \cdot B D \cdot C O}{A O \cdot B O-C O \cdot D O}}=\)
\[ =\tfrac{A C \cdot D O}{B D \cdot C O}+\tfrac{A O \cdot B O-C O \cdot D O}{B D \cdot C O}=\tfrac{A O \cdot D O+A O \cdot B O}{B D \cdot C O}=\tfrac{B D \cdot A O}{B D \cdot C O}=\tfrac{A O}{O C} \] which completes the proof.
Theorem 3 (Reformulation of Cevas theorem).
Let \(C_{1}\) and \(B_{1}\) be the points on the sides \(A B\) and \(C D\) of a triangle \(\triangle A B C\)
such that \(C C_{1}\) and \(B B_{1}\) meet at \(O\). Let \(A_{1}\) be the point on the side \(B C\). Line \(A A_{1}\) passes through the point \(O\) if and only if
\[ \tfrac{B A_{1}}{A_{1} C} \cdot \tfrac{C C_{1}}{B B_{1}} \cdot \tfrac{B_{1} O}{C_{1} O}=1 \]
Proof:
Using our triangle it suffices to prove \[ \tfrac{A Z}{B Z} \cdot \tfrac{B D}{A C} \cdot \tfrac{C O}{D O}=1 \]
Since \(Q K\) is a midline of triangle \(\triangle B O Z\), and \(K P\) is a midline of triangle \(\triangle A Z O\), then we have
\[ \begin{aligned} & \tfrac{A Z}{B Z} \cdot \tfrac{B D}{A C} \cdot \tfrac{C O}{D O}=\tfrac{P K}{Q K} \cdot \tfrac{B D}{A C} \cdot \tfrac{C O}{D O}= \\ & \tfrac{\tfrac{A C \cdot A B \cdot D O}{2(B D \cdot C O+A C \cdot D O}}{\tfrac{B D \cdot A B \cdot C O}{2(B D \cdot C O+A C \cdot D O)}} \cdot \tfrac{B D}{A C} \cdot \tfrac{C O}{D O}=\tfrac{A C \cdot D O}{B D \cdot C O} \cdot \tfrac{B D}{A C} \cdot \tfrac{C O}{D O}=1 \end{aligned} \] which completes the proof.
Corolary 4 (Stewart’s theorem)
Let \(\triangle A B C\) be a given triangle, and let D be a point on the segment \(B C\). Then
\[ A D^{2}=\tfrac{A C^{2} \cdot B D+B C^{2} \cdot A D-B C \cdot A D \cdot B D}{B C} \]
Proof:
Consider the point \(C\) and the triangle \(\triangle A B G\). t suffices to prove
\[ A C^{2}=\tfrac{A B^{2} \cdot C G+A G^{2} \cdot B C-B G \cdot B C \cdot G C}{B G} \]
Using the identities we have
\(\tfrac{A B^{2} \cdot C G+A G^{2} \cdot B C-B G \cdot B C \cdot G C}{B G}=\tfrac{A B^{2} \cdot 2 V R+4 V P^{2} \cdot B C-2 V Q \cdot B C \cdot 2 V R}{2 Q V}=\)
\(=A B^{2} \cdot \tfrac{A C \cdot D O}{B D \cdot A O}+\tfrac{A D \cdot A C \cdot B O}{A O \cdot B O-C O \cdot D O} \cdot \tfrac{A D \cdot A C \cdot B O}{B D \cdot A O}-\tfrac{B C^{2} \cdot A C \cdot D O}{A O \cdot B O-C O \cdot D O}=\)
\(=A C^{2}\left[\tfrac{A B^{2} \cdot D O}{B D \cdot A C \cdot A O}+\tfrac{A D^{2} \cdot B O^{2}}{B D \cdot A O(A O \cdot B O-C O \cdot D O)}-\tfrac{B C^{2} \cdot D O}{A C(A O \cdot B O-C O \cdot D O)}\right]=\)
\(=A C^{2} \cdot \tfrac{A B^{2} \cdot A O \cdot B O \cdot D O-A B^{2} \cdot C O \cdot D O^{2}+A D^{2} \cdot B O^{2} \cdot A C-B C^{2} \cdot B D \cdot A O \cdot D O}{B D \cdot A C \cdot A O \cdot(A O \cdot B O-C O \cdot D O)}=\)
\(=A C^{2} \cdot \tfrac{A B^{2} \cdot A O \cdot B O \cdot D O-A B^{2} \cdot C O \cdot D O^{2}+A D^{2} \cdot B O^{2} \cdot A C-B C^{2} \cdot B D \cdot A O \cdot D O}{B D \cdot A C \cdot A O \cdot(A O \cdot B O-C O \cdot D O)}\)
Now we will use some known identities for the quadrilateral:
\[ \begin{aligned} & \tfrac{A D^{2}-A O^{2}-D O^{2}}{A O \cdot D O}=\tfrac{A O^{2}+B O^{2}-A B^{2}}{A O \cdot B O} \Rightarrow \\ & A D^{2}=\tfrac{B O \cdot D O \cdot B D+A O^{2} \cdot B D-A B^{2} \cdot D O}{B O} \end{aligned} \] and
\[ \begin{aligned} & \tfrac{B C^{2}-B O^{2}-C O^{2}}{B O \cdot C O}=\tfrac{A O^{2}+B O^{2}-A B^{2}}{A O \cdot B O} \Rightarrow \\ & B C^{2}=\tfrac{A O \cdot C O \cdot A C+B O^{2} \cdot A C-A B^{2} \cdot C O}{A O} \end{aligned} \]
Using those identities we have \[ \begin{gathered} A B^{2} \cdot A O \cdot B O \cdot D O-A B^{2} \cdot C O \cdot D O^{2}+A D^{2} \cdot B O^{2} \cdot A C-B C^{2} \cdot B D \cdot A O \cdot D O= \\ A B^{2} \cdot A O \cdot B O \cdot D O-A B^{2} \cdot C O \cdot D O^{2}+\tfrac{B O \cdot D O \cdot B D+A O^{2} \cdot B D-A B^{2} \cdot D O}{B O} \cdot B O^{2} \cdot A C \\ -\tfrac{A O \cdot C O \cdot A C+B O^{2} \cdot A C-A B^{2} \cdot C O}{A O} \cdot B D \cdot A O \cdot D O= \\ A B^{2} \cdot A O \cdot B O \cdot D O-A B^{2} \cdot C O \cdot D O^{2}+B O^{2} \cdot D O \cdot B D \cdot A C+A O^{2} \cdot B O \cdot B D \cdot A C- \end{gathered} \] \[ \begin{aligned} & A B^{2} \cdot D O \cdot B O \cdot A C-A O \cdot C O \cdot A C \cdot B D \cdot D O-B O^{2} \cdot A C \cdot B D \cdot D O+A B^{2} \cdot C O \cdot B D \cdot D O \\ & =A B^{2}[B O \cdot D O(A O-A C)+C O \cdot D O(B D-D O)]+B D \cdot A C \cdot A O(A O \cdot B O-C O \cdot D O)= \end{aligned} \] \(A B^{2} \cdot(-B O \cdot D O \cdot C O+C O \cdot D O \cdot B O)+B D \cdot A C \cdot A O(A O \cdot B O-C O \cdot D O)=\) \[ B D \cdot A C \cdot A O \cdot(A O \cdot B O-C O \cdot D O) \]
So we have \[ \begin{aligned} & A C^{2} \cdot \tfrac{A B^{2} \cdot A O \cdot B O \cdot D O-A B^{2} \cdot C O \cdot D O^{2}+A D^{2} \cdot B O^{2} \cdot A C-B C^{2} \cdot B D \cdot A O \cdot D O}{B D \cdot A C \cdot A O \cdot(A O \cdot B O-C O \cdot D O)}= \\ & =A C^{2} \cdot \tfrac{B D \cdot A C \cdot A O \cdot(A O \cdot B O-C O \cdot D O)}{B D \cdot A C \cdot A O \cdot(A O \cdot B O-C O \cdot D O)}=A C^{2} \end{aligned} \] Which completes the proof.
REFERENCES
Akoshin, A.V, Zaslavski, A.A., 2007. Geometric properties of second order curves. MINMO Moscow.
Klamkin, M. S., February 1992. Simultaneous Generalization of the Theorems of Ceva and Menelaus. Mathematics Magazine, 65(1), 48 – 52.
Prasolov, V. Problems in plane and solid geometry. euclid.ucc.ie › mathenr › IMOTraining › planegeo.
Silvester, J.R., Mar. 2006. Extensions of a Theorem of Van Aubel Journal Article. The Mathematical Gazette, 90(517), 2 – 12.