**Angles**

Let’s prove that** sum of angles in a triangle is always 180 ^{0}**

Looking on a figure above and remembering what we learn about parallel lines we can tell:

If the line is parallel to the opposite side of the triangle. The anles marked** a** are equal, and so ar the angles marked

*.*

**c**We also know that angles that make up a straight line have a sum of **180 ^{0}** , so

**+**

*a*

*b + c =*180^{0}The angles inside the triangle are also ** a, b and c**.

**Сonsequence**

Because of any polygon with** n** sides can be divided into** n-2** triangles the **sum of the anles in any polygon with n sides is 180(n-2) ^{0}**

**Angle-Side relationship in Triangles**

Largest angle of a triangle is always across from the longest side, and vice versa. Likewise, the smallest angle is always across from the shortest side

An isosceles triangle is a triangle with two equal sides. If two sides in triangle are equal, then the angles across from those sides are equal, too, and vice versa.

**The triangle inequality**

The sum of any two sides of triangle is always greater than the third side. This mean that the length of any side of a triangle must be between the sum and the difference of the other two sides.

12-10 < AB < 12+10

2 < AB < 22

**The External anlge theorem**

The extended side of a triangle forms an external angle with the adjacent side. The external angle of a triangle is equal to the sum of the two “remote interior” angles.

Notice that this follows from our angle theorem:

* a + b + x = 180* and

*, therefore,*

**c + x = 180**

**a + b =c**** Problem 1**

In the figure above, if **AB = BD**, then what would be **x** ?

** Solution**

So if **AB = BD** then **∠BAD = ∠BDA**, in the triangle ABD : **∠BAD + ∠BDA** +50 = 180 therefore **∠BAD + ∠BDA** = 130 2***∠BAD** = 130, so **∠BAD** = 65

In triangle ACD : **∠A + ∠C +90 = 180 so **

**∠A +**, remembering that

**x = 90****∠A**will get

**= 65**

**:****x= 25**

**Problem 2**

In the figure above, if AD = DB = DC, then what would be x+y ?

** Solution**

In ∆ ABD : AD = DB ⤇ **∠**BAD = **∠**ABD = x, using that sum of anlges in triangle = 180, will get: 2x + 100 = 180 ⤇ 2x = 80 ⤇ x= 40;

Analogically, in ∆ BDC : BD = DC ⤇ **∠**BCD = **∠**DBC = y, **∠**BDC= 180 – 100 = 80, 2y + 80 = 180 ⤇ 2y = 100 ⤇ y = 50

So x + y = 50 +40 = 90.