How do you prove each of the following theorems using either a two-column paragraph or flow chart proof? triangle sum theorem-

triangle inequality theorem-

isosceles triangle theorem-

converse of the isosceles triangle theorem-

midsegment of a triangle theorem-

concurrency of medians theorem-

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Step-by-step explanation:

1) The Triangle Sum Theorem states that the sum of the angles in a triangle = 180°

2) The triangle inequality theorem states that the sum of any two sides of a triangle is larger than the third side

3) Isosceles triangle theorem states that the angles opposite the equal sides of an isosceles triangle are congruent

4) Converse of the Isosceles theorem states that the sides opposite the equal angles of an isosceles triangle are congruent

5) Midsegment of a triangle theorem states that the midsegment of two sides of a triangle is equal to half of the side it is parallel to

6) Concurrency of medians theorem states that the medians of a triangle intersect at a point within the triangle

Step-by-step explanation:

1) The Triangle Sum Theorem states that the sum of the angles in a triangle = 180°

Proof: To draw a triangle ABC starting from the point A we move 180° - ∠A to get to ∠B

From ∠B we turn 180° - ∠B to get to ∠C and from ∠C we turn 180° - ∠C to get back to A we therefore have turned 360° to get to A which gives;

180° - ∠A + 180° - ∠B + 180° - ∠C = 360°

Hence;

- ∠A - ∠B  - ∠C = 360° - (180°+ 180°+ 180°) = -180°

-(∠A + ∠B  + ∠C) = -180°

∴ ∠A + ∠B  + ∠C = 180°

2) The triangle inequality theorem states that the sum of any two sides of a triangle is larger than the third side

Proof: Given ΔABC with height h from B to D along AC, then

AC = AB×cos∠A + CB×cos∠C

Since ∠A and ∠C are < 90 the cos∠A and cos∠C are < 1

∴ AC < AB + CB

3) Isosceles triangle theorem

Where we have an isosceles triangle ΔABC with AB = CB, we have by sine rule;

Therefore;

sin(C) = sin(A) hence ∠A = ∠C

4) Converse of the Isosceles theorem

Where we have an isosceles triangle ΔABC with ∠A = ∠C, we have by sine rule;

Therefore;

sin(C) = sin(A) hence AB = CB

5) Midsegment of a triangle theorem states that the midsegment of two sides of a triangle is equal to half of the side it is parallel to

Given triangle ABC with midsegment at DF between BA and BC respectively, we have;

in ΔABC and ΔADF

∠A ≅ ∠A

BA = 2 × DA, BC = 2 × FA

Hence;

ΔABC ~ ΔADF (SAS similarity)

Therefore,

BA/DA = BC/FA = DF/AC = 2

Hence AC = 2×DF

6) Concurrency of Medians Theorem

By Ceva's theorem we have that the point of intersection of the segments from the angles in ΔABC is concurrent when the result of multiplying ratio the ratios of the segment formed on each of the triangle = 1

Since the medians bisect the segment AB into AZ + ZB

BC into BX + XB

AC into AY + YC

Where:

AZ = ZB

BX = XB

AY = YC

We have;

AZ/ZB = BX/XB = AY/YC = 1

∴ AZ/ZB × BX/XB × AY/YC = 1 and the median segments AX, BY, and CZ are concurrent (meet at point within the triangle).

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