Parmenides51 erxmer labis777 Cucumber Basil Athanasiadis Aerakis Aristides ATHANASIADIS Harry Alexi

Let angle ABC = i. Then angle ACB = pi / 2-i. Also angle ABS = angle CBQ = [p (angle ABC)] / 2 = (n-i) / 2 = n / 2-i / 2 angle BCQ = angle ACP = [p (angle ACB)] / 2 = [p (n / 2-i)] / 2 = (p-p / 2 + i) / 2 = (n / 2 + i) / 2 = pi / 4 + i / 2 and angle CAP = angle BAS = pi / 4. So angle BQC = p (angle CBQ) - (angle BCQ) = p (n / 2-i / 2) - (pi / 4 + i / 2) = pi-pi / 2 + i / 2-n / 4-i / 2 = pi / 2-pi / 4 = pi / 4. Therefore triangle ABS is similar to the BCQ, because it has two equal angles, whereby angle ASB = angle BCQ = pi / 4 + i / 2. Now brings the bisectors of triangle ABC which intersect jva at D. It DAP = angle angle angle DBQ = DCP = p / 2. The quadrilateral ADCP, BDCQ is recordable in a circle, because (angle DAP) + (angle DCP) = (angle DBQ) + (angle DCQ) = n / 2 + n / 2 = n. Therefore angle CDP = angle CAP = pi / 4 and angle BDC = p (angle BQC) = p-p / 4 = 3p / 4, so (angle CDP) + (angle BDC) = pi / 4 + 3p / 4 = n, so the points B, D, P are collinear and BP is height triangle PQS. Suppose that AC, BP intersect at E. In the triangle jva ADE is angle ADE = angle ACP = angle ASB = pi / 4 + i / 2 and angle DAE = angle BAS = pi / 4, so this is like triangles and ABS BCQ. Since D bring vertical DF and DG on AB and AC respectively, so DF = DG and AFDG rectangle is a square. So DF = DG = AF. The rectangular perimeter of triangle ABC is 2t = (AB) + (AC) + (BC) = 4 + 3 + 5 = 12, therefore t = 6 and (AF) = (DG) = t (BC) = 6-5 = 1. Bisectors theorem is now applied to the triangle: (AE) / (CE) = (AB) / (BC), or (AE) / [(AE) + (CE)] = (AB) / [(AB) + (BC )], or (AE) / (AC) = (AB) / [(AB) + (BC)], or (AE) / 3 = 4 / (4 + 5), or (AE) / 3 = 4 / 9 or (AE) = 3 * 4/9 = 4/3. Also, (BE) ^ 2 = (AB) ^ 2 + (AE) ^ 2 = 4 ^ 2 + (4/3) ^ 2 = 16 + 16/9 = (144 + 16) / 9 = 160/9, so (BE) = 4 * 10 ^ (1/2) / 3. From the similarity of triangles ABE DEG and get: (DE) / (DG) = (BE) / (AB), or (DE) / 1 = [4 * 10 ^ (1/2) / 3] / 4, or (DE) = 10 ^ (1/2) / 3. Eventually that (BQ) / (AE) = (BC) / (DE), or (BQ) / (4/3) = 5 / [10 ^ (1/2) / 3], or (BQ) = ( 4.3) * [15/10 ^ (1/2)] jva = 20/10 ^ (1/2) = 20 * 10 ^ (1/2) / 10 = 2 * 10 ^ (1/2) and ( BS) / (DE) = (AB) / (AE), or (BS) / [10 ^ (1/2) / 3] = 4 / (4/3), or (BS) = 3 * [10 ^ (1/2) / 3] = 10 ^ (1/2), so (QS) = (BQ) + (BS) = 2 * 10 ^ (1/2) + 10 ^ (1/2) = 3 * 10 ^ (1/2). The right triangle is isosceles BPQ because angle BQP = pi / 4, so (BP) = (BQ) = 2 * 10 ^ (1/2), whereby the area of the triangle is PQS: (PQS) = (1 / 2) * (QS) * (BP) = (1/2) * 3 * 10 ^ (1/2) * 2 * 10 ^ (1/2) = 3 * 10 = 30. Delete jva Reply
Parmenides51 erxmer labis777 Cucumber Basil Athanasiadis Aerakis Aristides ATHANASIADIS Harry Alexiou Ioannis Nikolaos Anastasopoulos? Stratis Aplakidis Giannis Antoniou Argirakis Atmatzidis Dimitrios Athanasios Vasakos Thomas Vassilas Nikolaos Georgakopoulos Ventistas George Elias Gidiaris Dimitrios Dimitrios Glenis Spiros Ginis Glavas Isidore Dodoneus EDUCATION Damian John Devetzis Jordan Dermitzoglou Diakoumakos Chris Duke George Manos Drougas Athanasios Dortsios Constantine Mandoulides Marios Eleftheriadis Eleftherios Baptist Zantaridis Zahariades Nikolaos Dimitrios Zahariades Lazarus Constantine Zorbas Zorbas Konstantinos Themelis POSITIVE Euripides Theodoropoulou Rebecca Panayiotis Theodoropoulos Theologian St. Nicholas Iosifides Kakavas Basil Kalfopoulou Katerina Kanavou Christos Nikolaos Kantidakis Kapellidis Spyros Panagiotis Karagiannis John Karagiorgos Karakastanias Athanasios Karaferi George Spyros Kardamitsi Kardasis Chris Herd Karkaletsis Theologis Athanasios Anastasios Kotronis Kouvaras Pantelis Andreas Koulouris Kourempanas George Koutsandreas Koutsovasilis jva Gerasimos Kyriakopoulos, Konstantinos Koutras Stathis Antonios Kyriakopoulos Ioannis Konstantopoulos Elias Kollias Stavros Lipordezis Athanasios Louridas Sotiris Lygatsikas Zenon Lympikis jva Aristides METHODIKO Mangos Mangos Thanos Michael Mitalas John Constantine Maggel Malliako Konstantinos jva Manolopoulos Mantzolas George Michael Nicholas Mavrogiannis Mavrofridis Basil Black John Mihailoglou Stelios Michalopoulos Nick Moschopoulos Dimitrios Moulogiannis Leonidas Barlas Anastasios Birmpakos Bounakis Dimitrios Dimitrios Dimitrios Bouzas may Rudolph Miller Dimitrios Panagiotis N. Mytarellis., and A. Bereketis INTELLECT Nannos Michael Alexander Nicholas Nicolaides Mars Nikolakakis Evangelos Davos Athanasios Christos jva Nikolopoulos Ntrizos Dimitrios Perros HORIZONS George Spyros Panagiotopoulos Papavlasopoulos Sozon Papagrigorakis Miltos Vassilis Papadakis Ioannis PAPAZI Papandreopoulos John Patsimas tiles Dimitrios Dimitrios Poulos Andreas Protopapas jva Eleftherios Rapanos Nikolaos Konstantinos Raptis Rizos George Thomas Raikoftsalis Sarafakis Nikolaos Dimitrios Roumeliotis Semsiris Aristides Sheriff Constantine Skotidas Sotirios Skokos Anthony Spatharas spleen Nikolaos Spyropoulos Dimitrios Dimitrios Stavropoulos Angelos Stergiou Babis Stefanidis jva Syngelakis Basil Alexander Sotiropoulos Tabakos Nicholas Constantine Christos Jeju Tzourounis Fotis Tzoumas Michael Tourlas Leonidas Trimis Pantelis Trifon Paul Tsakalakos Takis Diamandis Tsakmakidou Abraham Tsekoura Tsilpiridis Matthew Tsoulfanidis Emil Tsopelas John Togias Miltiades Fellouris Argyris Fresh Efstathios jva Fragakis Frantzeskos jva Nicholas George Kell Tutoring