## Sunday, September 28, 2008

### Elearn Geometry Problem 185

See complete Problem 185 at:
www.gogeometry.com/problem/p185_trapezoid_angles_problem.htm

Trapezoid, Triangle and angles. Level: High School, SAT Prep, College geometry

1. x=10

2. Solution of problem 185.
Let BC = u and CD = v. The extensions of AB and DC meet at E.
We have ang(CBE) = ang(DAB) = 4x. Since ang(ACD) = 5x and ang(CAE) = x, then ang(AEC) = 4x, so triangles ADE and BEC are isosceles, with CE = BC = u and DE = AD = AB = u+v.
Take F on AB so that AF = v. Then BF = BC = u, and triangle CBF is isosceles. As ang(CBE) = 4x, then ang(BFC) = ang(BCF) = 2x. Therefore, in triangle AFC, ang(ACF) = x = ang(CAF), so CF = AF = v.
Triangle ABD is isosceles so ang(ABD) = 90º - 2x. If G is the intersection of BD and DF we have ang(BGF) = 90º. So BD is the angle bisector of CBF and DF = CD = AF = v. This means that CDF is equilateral.
Hence x + 5x = 60º and x = 10º.

1. Good work Nilton

Did u have another solution Antonio?

2. dude there is a mistake in your reply,,, POINT G is the intersection of BD and CF and not BD and DF,,, rest I guess is true

3. The answer is x=10 , the solution of the equation
sin(4x)/sin(3x) == sin(8x)/sin(5x)

1. Por que se cumple esa ecuacion?

2. In triangle ADC, by sinus law
Because BC//AD => angleBCD = angleCAC = 3x => angle BCD = 8x
In triangle BCD, by sinus law
BD/sin(8x) = CD/sin(DBC)
In triangle BDA
BD/sin(4x) = AB/sin(BDA) = AB/sin(DBC) (angle DBC = angle BDA)
Replace in this relation
BD/sin(8x) = CD/sin(DBC) => ADSin(4x)/(sin(8x)Sin(DBC) = CD/sin(DBC)
Thus
sin(8x)/sin(4x) = sin(5x)/sin(3x) OR sin(4x)/sin(3x) = sin(8x)/sin(5x)

So you can solve this equation

4. The angle x satisfies the equation sin4x/sin3x = sin8x/sin5x whose solution in (0,pi/4) is x = 0.17453292519943298 rad = 10º

through C, make CF perpendicular to BD, F on AB;
so DC=DF; BCFD is a kite.
through A, make AH perpendicular to BD, H on BD;
so AH//CF, angle BFC=BAH=2X; so angle ACF=BAC=X; so FA=FC;
angle FCD =8X-2X =6X =CFD; so angle BF=8X; (BCFD is a kite)
so triangle CDF is equilateral triangle with angle 6X, so X=10 degree.

Let AH perpendicular to BD, AH is bissector of BAD. That means angle BAH = angle HAD = 2x.
AD // BC => angle CBD = angle BDA.
But ABD = BDA (base angle of isoceles triangle).. BD is bissector of ABC.
Let CF perpendicular to BD. We have FBC and FDC isoceles triangles. Then FB = BC, FD = CD.
CF // AH (CF and AH perpendicular to BD) , then triangle AFC isoceles and AF = FC.
Angle FCD = angle FCA + angle ACD = x+ 5x = 6x. This give angle CFD = 6x, BFD = 6x+2x = 8x.
Angle AFD = 180degree - 8x and angle FDA = 180 - (angle FAD+angle AFD)
Angle FDA = 180 -(180-8x + 4x) = 4x = angle BAD.
So triangle AFD is isoceles and AF = FD.

Because AF = FC, FD = CD, AF = FD then triangle FDC is equilateral and all angle = 6x.
So sum of all angle = 6x + 6x + 6x = 18x = 180 yield x = 10 degrees.

6. Based on Nilton Lapa's solution - a simplified version

Mark F on AB such that BF = BC

Tr.s BFD & BCD are congruent SAS (since < ABD = < ADB = < CBD)

So CD = FD.....(1)
Hence < DFC = < FCD = 6x and hence < FDA = 6x + 2x - 4x = 4x

Therefore FC = FA = FD = CD by (1)

This yields Tr. DFC equilateral and 6x = 60 and x = 10

Sumith Peiris
Moratuwa
Sri Lanka