Question

Theorem 6.7 : The sum of the angles of a triangle is 180°
Answer it for 12 points
You have to prove the theorem​

Answer 1

Answer:

Theorem

If ABC is a triangle then <)ABC + <)BCA + <)CAB = 180 degrees.

Proof

Draw line a through points A and B. Draw line b through point C and parallel to line a.

Since lines a and b are parallel, <)BAC = <)B'CA and <)ABC = <)BCA'.

It is obvious that <)B'CA + <)ACB + <)BCA' = 180 degrees.

Thus <)ABC + <)BCA + <)CAB = 180 degrees.

Lemma

If ABCD is a quadrilateral and <)CAB = <)DCA then AB and DC are parallel.

Proof

Assume to the contrary that AB and DC are not parallel.

Draw a line trough A and B and draw a line trough D and C.

These lines are not parallel so they cross at one point. Call this point E.

Notice that <)AEC is greater than 0.

Since <)CAB = <)DCA, <)CAE + <)ACE = 180 degrees.

Hence <)AEC + <)CAE + <)ACE is greater than 180 degrees.

Contradiction. This completes the proof.

Definition

Two Triangles ABC and A'B'C' are congruent if and only if

|AB| = |A'B'|, |AC| = |A'C'|, |BC| = |B'C'| and,

<)ABC = <)A'B'C', <)BCA = <)B'C'A', <)CAB = <)C'A'B'.

Answer 2

given: A and ABC

To prove: <A+ <B+ <C= 180°

const: draw a line xy passes through A which is to parallel to BC

proof: XY//BC ( by constructing)

<1=<B( alternate interior angles)

<2=<C( alternate interior angles)

but <1+<A+<2=180° ( straight line)

<B+<A+<C=180°

hence proved