Show that the orthocenter of ABC lies on P &#x2032; </msup> Q &#x203

uto2rimxrs50

uto2rimxrs50

Answered question

2022-05-08

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

Answer & Explanation

Mollie Roberts

Mollie Roberts

Beginner2022-05-09Added 21 answers

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

hetriamhageh6k20

Beginner2022-05-10Added 5 answers

Step 1
Suppose P Q intersect altitude B B on a point lie K. Draw a circle on side AC as it'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 C
That 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.

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