Geometry Problem. Post your solution in the comment box below.

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

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## Saturday, September 3, 2016

### Geometry Problem 1255: Right Triangle, Angle, 90 Degrees, Perpendicular, Collinear Points

Labels:
angle,
collinear,
perpendicular,
right triangle

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Problem 1255

ReplyDeleteSolution 1

In accordance with problem 848 in triangle BCD <BDC=26.5 so <BCD=90-26.5=63.5.

But accordance with problem 849 in triangle CDE <DEC=18.5 so <DCE=90-18.5=71.5.

Now <ACB+<BCD+<DCE=45+63.5+71.5=180.Therefore A,C, and E are collinear.

Solution 2

Let the points K,L on the side DE such that DK=KL=LE. If <DKC=α, <DLC=β and <DEC=θ

then in accordance with problem 363 in triangle DCE apply α+β+θ=90 or β+θ=45.

Is triangle BCD similar with triangle DCL(BC=2BD, DC=2DL) so <BCD=<DCL=90-β and

<DCE=90-θ. But <ACB+<BCD<DCE=45+(90-β)+(90-θ)=45+180-(β+θ)=45+180-45=180.

Therefore A,C, and E are collinear

APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE

Geometry solution:

ReplyDeletehttps://goo.gl/photos/sdw3M9H3Ktv2F8PVA

Let x=AB=BC and BD=2x

Let AC cut circumcircle of triangle BDC at F

Note that ADF is 45-45-90 triangle so

DF=AF=3x/√(2)

CF=AF-AC=3x/√(2) – x√2= x/√(2)

And CD/DE=CF/DF=1/3

Triangle DFC similar to EDC ( case SAS)

So angle (DCF)= angle (DCE)=> A, C and E are collinear.

Trigonometry solution:

Let u= ∠ (BCD) and v= ∠ (DCE)

We have ∠ (BCE)=u+v

So tan(∠ (BCE))= (tan(u)+tan(v))/(1- tan(u).tan(v))= (2+3)/(1-2.3)= -1

So ∠ (BCE=135 degrees and ∠ (ACE)= 45+135= 180 degrees

So A,C and E are collinear

Geometry Solution

ReplyDeleteDrop a perpendicular EF from E to AF.

Let AB = BC = p and CD = q so that BD = 2p and DE = 3q

Now triangles CBD and DFE are similar hence DE = 3p and FE = 6p since DE = 3DC.

So AF = 6p = FE hence < FAE = 45. But < FAC = 45

It thus follows that A,C,E are collinear

Sumith Peiris

Moratuwa

Sri Lanka

Trigonometry Solution

ReplyDeleteLet < BCD = m and let < DCE = n

So tan(m+n) = (2+3)/(1-2X3) = -1

Hence m+n = 135 degrees and the result follows

Sumith Peiris

Moratuwa

Sri Lanka

Let AB = BC = 1 => AC = Sqrt(2)

ReplyDeleteBD = 2 => DC = Sqrt(5)

Join CE and form the Right triangle CDE

CD = Sqrt(5)

DE = 3*Sqrt(5)

CE = 5*Sqrt(2)

Drop a _|_ from D to EC that meets at F, DF = 3/Sqrt(2)

Area of ADC = 0.5*AD*BC = 0.5*3*1 --------(1)

Area of DCE = 0.5*DC*DE = 0.5*Sqrt(5)*3*Sqrt(5) = 0.5*5*3 -------------(2)

Adding (1) & (2) = 9 Sq.units ---------(3)

Let us assume A,C & E are collinear

Join AE and form the triangle ADE.

Area of ADE = 0.5*AE*DF = 0.5*6*Sqrt(2)*3/Sqrt(2) = 9 Sq.units ------------(4)

Since (3) = (4), our assumption is true and A,C & E are collinear