## Saturday, September 3, 2016

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

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to view more details of problem 1255.

1. Problem 1255
Solution 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

2. Geometry solution:
https://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

3. Geometry Solution

Drop 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

4. Trigonometry Solution

Let < 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

5. Let AB = BC = 1 => AC = Sqrt(2)
BD = 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)