<prompt> Katika nafasi ya kuratibu, A 1,2,3, B 5,3,1, na C 3,4,5. Tafuta kiini cha pembetatu ABC.  <response>: Acha D,E,F iwe miguu ya miinuko kutoka A,B, na C, mtawalia. Acha H iwe kituo cha orthocenter.asyukubwa wa kitengo 0.6 cm;jozi A, B, C, D, E, F, H;A 2,5;B 0,0;C 8,0;D A tafakariB,CA2;E B tafakariC,AB2;F C tafakariA,BC2;H kiendeleziA,D,B,E;kuchoraA--B--C--mzunguko;kuchoraA--D;kuchoraB--E;kuchoraC--F;leboA, A, N;leboB, B, SW;leboC, C, SE;leboD, D, S;leboE, E, NE;leboF, F, NW;leboH, H, SE, UnFill;asyKumbuka hilooverrightarrowBA anzapmatrix 1 - 5 2 - 3 3 - 1 mwishopmatrix anzapmatrix -4 -1 2 mwishopmatrix quad textna quad overrightarrowBC anzapmatrix 3 - 5 4 - 3 5 - 1 mwishopmatrix anzapmatrix -2 1 4 endpmatrix.Kisha makadirio ya overrightarrowBA kwenye overrightarrowBC nioverrightarrowBD fracoverrightarrowAB cdot overrightarrowBCoverrightarrowBC cdot overrightarrowBC overrightarrowBC fracbegin pmatrix -4 -1 2 mwishopmatrix cdot anzapmatrix -2 1 4 mwishopmatrixleft anzapmatrix -4 -1 2 mwishopmatrix kulia kushoto anzapmatrix -2 1 4 mwishopmatrix kulia anzapmatrix -2 1 4 mwishopmatrix frac1521 anzapmatrix -2 1 4 mwishopmatrix anzapmatrix -107 57 207 mwishopmatrix.Inafuata kwambaoverrightarrowAD overrightarrowAB overrightarrowBD anzapmatrix 4 1 -2 mwishopmatrix anzapmatrix -107 57 207 endpmatrix anzapmatrix 187 127 67 endpmatrix.Kumbuka kwamba hii ni sawia na begin pmatrix 3 2 1 mwishopmatrix. Kwa hivyo, mstari AD unaweza kuainishwa naanzapmatrix 1 3t 2 2t 3 t mwishopmatrix.Kuweka hii kuwa H, tunapataoverrightarrowCH anzapmatrix 1 3t 2 2t 3 t mwishopmatrix - anzapmatrix 3 4 5 mwishopmatrix anzapmatrix -2 3t -2 2t -2 t mwishopmatrix.Vekta hii itakuwa orthogonal kwa overrightarrowBA, hivyoanzapmatrix -2 3t -2 2t -2 t mwishopmatrix cdoti anzapmatrix -4 -1 2 mwishopmatrix 0.Kisha -2 3t-4 -2 2t-1 -2 t2 0. Kutatua, tunapata t frac 12. Kwa hiyo, H left frac52, 3, frac72 kulia. 