Question

Suppose A*B*A′ holds(B is between A and A')  and D ∈ Int(∠ABC). Prove that C ∈ Int(∠A′BD).

(a) Prove that C ∈ H(D,line A′B).

(b) Prove that C ∈ H (A′, line←→BD). Use point A.

(c) Deduce that C∈Int (∠A′BD).

Answer 1

a).

If A and D are points on the same side of a line ←→BC and the line ←→BA is parallel to the line ←→CD

then m (∠ABC) + m (∠BCD) = 180

To obtain a contradiction, suppose that ←→CD then m (∠ABC) + m (∠BCD) 6= 180.

Case 1:

Suppose m (∠ABC) + m (∠BCD) < 180.

Then, applying Euclid’s 5th Postulate, the lines ←→BA and ←→CD meet at a point P on the same side of ←→BC as A and D. But ←→BA and ←→CD are parallel,

so this is a contradiction.

Case 2:

Suppose m (∠ABC) + m (∠BCD) > 180. Let E be a point on ←→AB such that A − B − E. Similarly, let F be a point on ←→CD such that D − C − F.

Using the supplement postulate and the fact that ∠ABC and ∠CBE form a linear pair and that ∠DCB and ∠BCF also form a linear pair: m (∠ABC) +m (∠CBE) = 180, m (∠DCB) +m (∠BCF) = 180, and by assumption,

m (∠ABC) +m (∠BCD) > 180. Hence m (∠CBE) + m (∠BCF) < 180.

Again applying Euclid’s 5th Postulate, the lines ←→BA and ←→CD meet at a point P on the same side of ←→BC as E and F.

But ←→BA and −−→CD are parallel, so this is again a contradiction.

Thus

m (∠ABC) + m (∠BCD) = 180.

b)

Every distinct pair of parallel lines have a common perpendicular.

Suppose ←→AB is parallel to the line ←→CD (and suppose these lines are distinct).

Let E be a point in the opposite half plane of ←→AB as the point C (since ←→AB is parallel to the line ←→CD, C is not on the line ←→AB).

there is a unique perpendicular to ←→CD through the point E.

Let F be the point of intersection between this perpendicular and the line ←→CD.

Since E and C are in opposite half planes, by the Plane Separation Axiom, EF intersects ←→AB at some point G

assume A − G − B and C − F − D. Notice that we also have E − G − F and m (∠GF D) = 90.

Applying   ←→AB, ←→CD, and ←→EF, we have that m (∠GF D) + m (∠BGF) = 180. Hence m (∠BGF) = 90, and the line ←→EF is perpendicular to both ←→AB and ←→CD.

c)

The sum of the measures of any triangle is 180.

Consider a triangle △ABC.

Let ←→CE be the unique line through C parallel to the line ←→AB. Let F be a point such that F − C − E and G and H be points such that G − A − B − H.

Then, applying 12(c) to the parallel lines ←→AB and ←→CE and the transversal ←→AC, m (∠ACF) = m (∠CAB).

Similarly, applying 12(c) to the parallel lines ←→AB and ←→CE and the transversal ←→CB, m (∠ECB) = m (∠CBA).

Notice that ∠F CA and ∠ECA form a linear pair, so by the Supplement Postulate, m (∠F CA)+m (∠ECA) = 180.

Notice that since A and B are on the same side of ←→CE and we have A − B − H, then B ∈ int(∠F CA).

Thus m (∠F CA) = m (∠ACB) + m (∠F CB). Therefore, m (∠ECA) + m (∠ACB) + m (∠F CB) = 180. But then m (∠CAB) + m (∠ACB) + m (∠CBA) = 180. Hence the sum of the angles of △ABC is 180.