Euclidean Version of Pappus's theoremLet A, B, C, be points on a line l, and let A′, B′, C′ be points on a line m. Assume AC′∥A′C and B′C∥BC′. Show that AB′∥A′B.

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Euclidean Version of Pappus&#039;s theoremLet A, B, C, be points on a line l, and let A′, B′, C′ be points on a line m. Assume AC′∥A′C and B′C∥BC′. Show that AB′∥A′B.

heiftarab · Accepted Answer

Step 1Pappus&#039; theorem is a simple consequence of similarity of Euclidean triangles (in guise of the intercept theorem) and there&#039;s no need of introducing the circle:Let P be the point of intersection of l and m. All we need to do is to show that             P              A        ′                    P      B        =            P              B        ′                    P      A      .Step 2But by assumption we have             P              A        ′                    P      C        =            P              C        ′                    P      A       and             P      C              P      B        =            P              B        ′                    P              C        ′            . Multiplying the left hand sides and the right hand sides together, we get what we want.

Ledexadvanips · Accepted Answer

Step 1By using three cyclic quadrilaterals, we can establish the desired answer. First, let us prove an almost obvious lemma: Let ABCD be a cyclic quadrilateral, where AC is one of its diagonals. Let A&#039;C&#039; be parallel to AC and A&#039; on AD and C&#039; on extension of BC. Then A&#039;BC&#039;D is cyclic. The proof is quite clear:   ∠      A    ′    D  B  =  ∠  D      C    ′    B for ABCD is cyclic and A&#039;C&#039;∥AC.Step 2This lemma is used to show CB&#039;A&#039;D is cyclic, all notations as in the original thread. Hence   α  :=  ∠  C      B    ′    D  =  ∠  D      A    ′    C. Now we examine the following relations:   ∠  B      C    ′        A    ′    =  ∠  C      B    ′        C    ′    =  ∠  D      B    ′        C    ′    +  α  =  ∠  D  A      C    ′    +  α  =  ∠      A    ′    C  D  +  α  =  ∠  B  D      A    ′  . They lead to the conclusion that DBA&#039;C&#039; is cyclic. Stacked up with the cyclic quadrilateral AB&#039;C&#039;D with collinear sides, the relations   ∠  B      A    ′        C    ′    =  ∠  A  D      C    ′    =  ∠  P      B    ′    A are clear.This show AB′∥A′B

Let A, B, C, be points on a line l, and let A′, B′, C′ be points on a line m. Assume AC′∥A′C and B′C∥BC′. Show that AB′∥A′B.

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