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MA_775_DIABLO [31]
1 year ago
7

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

Mathematics
1 answer:
lilavasa [31]1 year ago
8 0

All the theorems are proved as follows.

<h3>What is a Triangle ?</h3>

A triangle is a polygon with three sides , three vertices and three angles.

1. The Triangle sum Theorem

According to the Triangle Sum Theorem, the sum of a triangle's angles equals 180 degrees.

To create a triangle ABC, starting at point A, move 180 degrees away from A to arrive at point B.

We turn 180 degrees from B to C and 180 degrees from C to return to A, giving a total turn of 360 degrees to arrive to A.

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

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

∠A + ∠B  + ∠C = 180°

(Hence Proved)

2. Isosceles Triangle Theorem

Considering an isosceles triangle ΔABC

with AB = AC, we have by sine rule;

\rm \dfrac{sinA}{BC} =  \dfrac{sinB}{AC} =  \dfrac{sinC}{AB}\\

as AB = AC

sin B = sin C

angle B = angle C

3.Converse of the Isosceles theorem

Consider an isosceles triangle ΔABC with ∠B= ∠C, we have by sine rule;

\rm \dfrac{sinA}{BC} =  \dfrac{sinB}{AC} =  \dfrac{sinC}{AB}\\

as  ∠B= ∠C ,

AB = AC

4. Midsegment of a triangle theorem

It states that the midsegment of two sides of a triangle is equal to (1/2)of the third side parallel to it.

Given triangle ABC with midsegment at D and F of AB and AC respectively, DF is parallel to BC

In ΔABC and ΔADF

∠A ≅ ∠A

BA = 2 × DA, BC = 2 × FA

Hence;

ΔABC ~ ΔADF (SAS similarity)

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

Hence AC = 2×DF

5.Concurrency of Medians Theorem

A median of a triangle is a segment whose end points are on vertex of the triangle and the middle point of the side ,the medians of a triangle are concurrent and  the point of intersection is inside the triangle known as Centroid .

Consider a triangle ABC , X,Y and Z are the midpoints of the sides

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

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).

To know more about Triangle

brainly.com/question/2773823

#SPJ1

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