# Thread: Help with simple geometry :D

1. ## Help with simple geometry :D

Hello, I'm new to math help forums. I hope you could help me.

I have a test tomorrow and I'm having trouble with a certain subject. I don't know what it's called in English, but it's those triangles that have at least one 90 degrees angle in them. It's pretty simple. Just basic triangle rules. But the tests are very tricky and I got this work that's supposed to prepare me but I can't solve anything! If you could help me it would make it a lot easier for me to figure out how to solve such problems in the future. Ok, so, here we go.

There's this triangle

Here's the info(I'm from Israel so forgive me if I don't hit all the terms :P)

Triangle ABC is a right triangle(AB perpendicular to BC)
BD is the midpoint of AC
AE = CE
AE perpendicular to BD

Calculate the angle C

I don't really know how to use the MATH tags so sorry if it looks a little bad. I tried solving it, but I can't figure out how I'm supposed to get angles here if the only angle I know it's size is B which is 90?

2. Originally Posted by thejack
Hello, I'm new to math help forums. I hope you could help me.

...
There's this triangle

Here's the info

Triangle ABC is a right triangle(AB perpendicular to BC)
BD is the midpoint of AC
AE = CE
AE perpendicular to BD

Calculate the angle C

I don't really know how to use the MATH tags so sorry if it looks a little bad. I tried solving it, but I can't figure out how I'm supposed to get angles here if the only angle I know it's size is B which is 90? <== that's wrong. See the highlighted part of the question.

1. I've modified your sketch a little bit (see attachment).

2. Since $\displaystyle |\overline{AE}|=|\overline{CE}|$ the triangle ACE is isosceles. Let x denote the value of the angle at C then the exterior angle of the triangle at E must have the value 2x.

3. Since $\displaystyle |\overline{BC}| = |\overline{BD}| = |\overline{BA}|$ (circle of Thales!) the triangle CBD is isosceles. Therefore the angle $\displaystyle \angle(DBC)$ has a value of x.

4. Take the greyed triangle. The sum of all interior angles of this triangle is:

$\displaystyle 2x+x+90^\circ=180^\circ$

Solve for x.

3. Originally Posted by earboth
1. I've modified your sketch a little bit (see attachment).

2. Since $\displaystyle |\overline{AE}|=|\overline{CE}|$ the triangle ACE is isosceles. Let x denote the value of the angle at C then the exterior angle of the triangle at E must have the value 2x.

3. Since $\displaystyle |\overline{BC}| = |\overline{BD}| = |\overline{BA}|$ (circle of Thales!) the triangle CBD is isosceles. Therefore the angle $\displaystyle \angle(DBC)$ has a value of x.

4. Take the greyed triangle. The sum of all interior angles of this triangle is:

$\displaystyle 2x+x+90^\circ=180^\circ$

Solve for x.
thanks

4. Hello earboth.
I Think your statement 3 needs correction

bjh

5. Just a typo by Earboth.

He meant $\displaystyle |\overline{DC}|=|\overline{DA}|=|\overline{DB}|$