# The hardest easy geometry problem.

#### Veho

##### The man who cried "Ni".
OP
Former Staff

Find the value of x, with mathematical proof. No advanced trigonometry allowed. No algebra needed. Here's all you need to know:

Lines and Angles: When two lines intersect, opposite angles are equal and the sum of adjacent angles is 180 degrees. When two parallel lines are intersected by a third line, the corresponding angles of the two intersections are equal.

Triangles: The sum of the interior angles of a triangle is 180 degrees. An isosceles triangle has two equal sides and the two angles opposite those sides are equal. An equilateral triangle has all sides equal and all angles equal. A right triangle has one angle equal to 90 degrees. Two triangles are called similar if they have the same angles (same shape). Two triangles are called congruent if they have the same angles and the same sides (same shape and size).

* Side-Angle-Side (SAS): Two triangles are congruent if a pair of corresponding sides and the included angle are equal.
* Side-Side-Side (SSS): Two triangles are congruent if their corresponding sides are equal.
* Angle-Side-Angle (ASA): Two triangles are congruent if a pair of corresponding angles and the included side are equal.
* Angle-Angle (AA): Two triangles are similar if a pair of corresponding angles are equal.

Have fun going insane.

Member
its less than 50

#### xcalibur

##### Gbatemp's Chocolate Bear
Member
easy,
for this triangle, lets name the middle point x

angle AXB is 50 degrees 180 - 130 = 50
that makes AXD 130 because 180 - 50 = 130
BXE is 130 also and DXE is 50

QUOTE said:
Lines and Angles: When two lines intersect, opposite angles are equal and the sum of adjacent angles is 180 degrees. When two parallel lines are intersected by a third line, the corresponding angles of the two intersections are equal.

angle XBE is 30 degrees to make triangle BXE a full 180 degrees
triangle BDE is a right angled triangle so that makes angle BDE 90 degrees
so that means 50 + 90 = 140 and that makes and DEB (x) = 40

my geometry is a bit rusty so this might be completely wrong

#### Veho

##### The man who cried "Ni".
OP
Former Staff
angle XBE is 30 degrees to make triangle BXE a full 180 degrees

Wrong. XBE is 20, says so on the picture.

#### xcalibur

##### Gbatemp's Chocolate Bear
Member
angle XBE is 30 degrees to make triangle BXE a full 180 degrees

Wrong. XBE is 20, says so on the picture.

I meant angle XEB
I'm sorry, when you do something longwinded you tend to get lost.
try doing trigonometric identities... those are loooooooong

#### Veho

##### The man who cried "Ni".
OP
Former Staff
Okay, but it's still not it.

triangle BDE is a right angled triangle so that makes angle BDE 90 degrees

Why do you think BDE is 90°? It looks like it, but there's no way to know that for sure.

Member
it's 20°

#### Veho

##### The man who cried "Ni".
OP
Former Staff
it's 20°
Prove it. Yeah, I know it's supposed to be 20°, but how do you get there? I need the procedure.

EDIT:

A picture drawn to scale.

BDE is 110°.

#### xcalibur

##### Gbatemp's Chocolate Bear
Member
Okay, but it's still not it.

triangle BDE is a right angled triangle so that makes angle BDE 90 degrees

Why do you think BDE is 90°? It looks like it, but there's no way to know that for sure.

DAMNIT... i was hoping i wouldnt have to round it the loong way but i guess ill have to now :'(
serves me for trying to take a shortcut

angle ACB is 20
angle BDC is 140
angle AEB is 50

and thats where im stuck...
i cant seem to figure our trianle CDE and i need to go now...

EDIT: i saw your post so i can get it now
BDE is 110
DXE is 50
that makes DEX 110 +50 = 160
180 - 160 = 20

howd you get BDE?

#### kellyan95

##### Banned!
Banned
Print -> use protractor

#### xcalibur

##### Gbatemp's Chocolate Bear
Member
Print -> use protractor
not to scale

EDIT: i guess it is...
that was a useless post

#### Veho

##### The man who cried "Ni".
OP
Former Staff
howd you get BDE?
Um...

I don't know. I can't prove it. I know it's 110, and x=20°, but I can only prove it graphically, and this problem asks for discrete proof.

#### shakirmoledina

##### Legend
Member
Some unknown angles are known but if cockroachman could explain how he got 20...
ADB = 40... AEb = 30.... EPB and DPA are 130 where P is central point and APB and DPE are 50

#### CockroachMan

##### Scribbling around GBATemp's kitchen.
Member
don't know how to prove it..

for some reason, the triangles B-Center-E and DCE are equal.. but I can't remember why now.. something to do with the lines crossing them..

been some time since I've done this kind of thing..

#### shakirmoledina

##### Legend
Member
umm just a note that here we dont use AAS since
* Angle-Side-Angle (AAs)
* Angle-Angle (AA)
as u can see above if 2 angles are equal why do u need to prove the side

#### FrEEz902

##### Well-Known Member
Member
Easy,

X=60 degrees

I'll scan my paper of ugly handwriting now

Edit: Oh wait, i screwed something up >_

#### TGBoy

##### Well-Known Member
Member
I have seen this before. Isnt this the World's Hardest Easy Geometry Problem.

I seen the solution to it.

#### ZAFDeltaForce

##### Specialist
Member
Easy,

X=60 degrees

I'll scan my paper of ugly handwriting now

Edit: Oh wait, i screwed something up >_

#### moozxy

##### hamboning
Member
Easy,

X=60 degrees

I'll scan my paper of ugly handwriting now

Edit: Oh wait, i screwed something up >_

#### FrEEz902

##### Well-Known Member
Member
Easy,

X=60 degrees

I'll scan my paper of ugly handwriting now

Edit: Oh wait, i screwed something up >__<

Either way it doesn't change much.

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