Online Geometry theorems, problems, solutions, and related topics.
See complete Problem 5 at:www.gogeometry.com/problem/problem005.htmTriangle, Angles, Congruence. Level: High School, SAT Prep, College geometryPost your solutions or ideas in the comments.
Triangle CDB is isosceles since BD=DC, therefore angle BCD=angle DBC=2x.Let point E be a point inside ABD triangle so that AED triangle is congruent to CDB triangle, where AE=ED=BD=DC and as it's given AD=BC. Note that both angles EAB and EAD are equal 2x.Angle ADB = 4x=>angle EDB=2x. Triangle EDB is isosceles since ED=BD.Angle BAE = angle BAD - angle EAD = 3x - 2x = x.Let's reflect point E over AB calling it F. Triangles AFB and AEB are congruent. AF=AE, FB=EB. Angle FAE = 2 * angle BAE = 2x.FAE and BDE are congruent isosceles triangles since AF=AE=DE=DB and angle FAE=angle EDB=2x=>FE=EB, but FB=EB => triangle FBE is equilateral => angle FBE=60=>AEB=30.Angle EBD = (180 - angle EDB) / 2 = 90 - x.Angle ABC = 180 - angle BAC - angle BCA = 180 - 5x.But angle ABC = angle ABE + angle EBD + angle DBC, or180 - 5x = 30 + (90 - x) + 2x=> 180 - 5x = 120 +x => x=10
"Note that both angles EAB and EAD are equal 2x"What? How come? If you subtract <EAD=2x from <BAD=3x, the remainder <BAE=x, not 2x.
Video solution (Spanish version) by Eder Contreras and Cristian Baeza at Geometry problem 5. Thanks Eder and Cristian.
Draw an angular bisector of BDA.Draw DAE=2x, so that the angular bisector of BDA meets it at E.So, AE=ED (EAD=EDA=2x)But ∆EAD and ∆BDC are congruent under A.A.S case)So, ED=BDNow, Draw a Circle having the center D and Radius DCObviously it passes through E, B and CLet AC and the circle meet at GSo, GD=BD (Radius)Now connect BGH be the point of intersection of BG and DESince ∆BDG is an Isosceles DH is perpendicular to BGSo, HGD=90-2xLet F be the reflex of E over ABAs AF=AE, FAB=EAB=x, AB=AB∆AFB and ∆ AEB are congruentSo FB=EBAlso, ∆AFE and ∆BDE are congruent (BD=AF, BDE=FAE=2x, DE=AE)So, BE=FEEventually EB=EF=BFSo, ∆BFE is an equilateralTherefore FBE=600Since FE and AB are perpendicular to each otherABE=300Since, Ang.ABC=ABE+EBD+DBC,ABC=300+900-x+2xABC=1200+xBut GBC=90Deg as it is an angle on semicircle (Note that GC is a diameter)So, ABG=120+x-90ABG=30+xSince BGD=BAG+ABG900-2x=3x+300+x600=6xSo, x=100Note: Since ABD = 120-x the proof for the “PROBLEM 4 is inbuilt”
Read "x = 10 degrees" instead "x=100"
Extend BD to E such that BD = DE Verify BCE is a right angle; Angle BDA = 4x; ABD = Pi - 7xNow cos 2x = BC / BE = BC / 2BD=> 2 cos 2x = BC / BD = AD / BD = sin (pi - 7x) / sin3x = sin 7x / sin 3x=> 2 sin 3x cos 2x = sin 7x=> sin 5x + sin x = sin 7x=> sin 7x - sin 5x = sin x=> 2 cos 6x sin x = sin x=> cos 6x = 1/2=> 6x = pi/3=> x = pi/18 = 10 deg
Dear Antonio - is there a proof using properties of 80-20-80 triangles?