Show that the orthocenter of ABC lies on P ′ Q ′ , where P ′ , Q ′ , R ′ are the symmetric points of M to this sides of the triangle, M on circumscribed

Question

Show that the orthocenter of ABC lies on       P    ′        Q    ′   , where       P    ′    ,      Q    ′    ,      R    ′   are the symmetric points of M to this sides of the triangle, M on circumscribed

Mollie Roberts · Accepted Answer

Step 1      P    ′   and       Q    ′   are the symmetric points of M w.r.t. AB, BC. Let D be the intersection of the circle and the perpendicular line from A to BC. So H and D are symmetric with respect to BC (you could prove it yourself). Let S be the symmetric point of       P    ′   w.r.t. BC. So, to prove       P    ′    ,      Q    ′    ,  H are collinear, we need to prove M, D, S are collinear.We have   ∠  C  B  D  =  ∠  C  A  D  =      90    ∘    −  ∠  A  C  B , so   ∠  A  M  D  =  ∠  A  B  D  =  ∠  A  B  C  +      90    ∘    −  ∠  A  C  B . Also,   ∠  A  M  B  =  ∠  A  C  B , and since we have   M  B  =      P    ′    B  =  S  B ,   ∠  M  B  S  =  ∠  C  B  S  −  ∠  C  B  M  =  ∠  C  B  P  −  ∠  C  B  M  =  ∠  A  B  C  +  ∠  A  B  P  −  ∠  C  B  M  =  ∠  A  B  C  +  ∠  A  B  M  −  ∠  C  B  M  =  2  ∠  A  B  C . So, we have   ∠  B  M  S  =      90    ∘    −  ∠  A  B  C . Therefore,   ∠  D  M  S  =  ∠  D  M  A  +  ∠  A  M  B  +  ∠  B  M  S  =  ∠  A  B  C  +      90    ∘    −  ∠  A  C  B  +  ∠  A  C  B  +      90    ∘    −  ∠  A  B  C  =      190    ∘   . So D,M,S are collinear.

hetriamhageh6k20 · Accepted Answer

Step 1Suppose       P    ′        Q    ′   intersect altitude   B      B    ′   on a point lie K. Draw a circle on side AC as it&#039;s diameter, it intersect   B      B    ′   at point F. Connect C to K and extend it to meet circle on AC at point L. Extend altitude   B      B    ′   to meet this circle at point D. Draw a line from L parallel with BD to meet side AC at point E.In triangle       B    ′    K  C we have:  ∠      B    ′    K  C  =                              L          F                          ⌢                    +                        C          D                          ⌢                      2  In triangle LAC we have:  ∠  L  A  C  =                    L        C                    ⌢              2  But:            F      C              ⌢        =            C      D              ⌢      Therefore:  ∠      B    ′    K  C  =  ∠  L  A  C  ⇒  △      B    ′    K  C  ∼  △  A  L  CThat means   ∠  A  L  C  =      90    o   .The extension of AL meets altitude   B      B    ′   at B because triangles ALE and   A  B      B    ′   are similar. That means CL is an altitude from C on side AB. In this way K is coincident with orthocenter H.

Show that the orthocenter of ABC lies on P ′ </msup> Q &#x203

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