## Sunday, February 3, 2013

### Problem 851: Triangle, Triple side, Angles, 90 Degrees, Special Right triangle, Sides ratio 1:3

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 851.

1. Bx perpendicular to the AB
Bx and AC meet at point D. Then , angle DBC=angle A , and the triangles ABC ,DBC are similar.
BD/AB=BC/AC=1/3. So ,AB=3AD .Therefore (problem 849) , angle A=18.5 degrees

2. Locate point D on AC such that AB ⊥ BD.
Triangle BDC similar to triangle ABC ( case AA)
so BC/AC=BD/AB=1/3
so angle A= 18.5 per problem 849

3. Construim AD ⊥ AC si luam punctul D astfel incat AD =a , D si B sa fie de o parte si de alta a dreptei AC.Vom obtine
trapezul isoscel ABCD (A= 18.5 per problem 849

4. Draw line from B, meeting AC in D, with angle(CBD) = 90°.
Then angle(ABD) = angle(A), so triangle ABD is isosceles.
AD = x, then BD = x and CD = 3a - x.
Pythagoras in triangle BCD: (3a-x)² = x² + a², giving x = 4/3·a.
BC = a, BD = x = 4/3·a, CD = 3a-x = 5/3·a.
So triangle BCD is the well-known 3-4-5-triangle!
Angle(BDC) = 2·angle(A)
From the 3-4-5-triangle the angle is well-known, being (approx.) 36,87°.
So angle(A) = 36,87° / 2 = 18,4° (approx.).