Question

What is Pythagoras theorem and mid point theorm

Answer 1

mid point theorem: The line segment connecting the midpoints of  two sides of a triangle is parallel to the third side and is congruent to one half of the third side.

Pythagoras theorem:In a right angled triangle:
the square of the hypotenuse is equal to
the sum of the squares of the other two sides.
ab^2 + bc^2=ac^2

Answer 2

Pythagoras Theorem Statement

Pythagoras theorem states that “In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides“. The sides of this triangles have been named as Perpendicular, Base and Hypotenuse. Here, the hypotenuse is the longest side, as it is opposite to the angle 90°. The sides of a right triangle (say x, y and z) which has positive integer values, when squared are put into an equation, also called a Pythagorean triple.

and

MidPoint Theorem Statement

The midpoint theorem states that “The line segment in a triangle joining the midpoint of two sides of the triangle is said to be parallel to its third side and is also half of the length of the third side.”

Pythagoras Theorem Proof

Given: A right-angled triangle ABC.

To Prove- AC² = AB² + BC²

Proof: First, we have to drop a perpendicular BD onto the side AC

We know, △ADB ~ △ABC

Therefore,

AD AB

----- = -----

AB AC

(Condition for similarity)

Or, AB² = AD × AC …………………..……..(1)

Also, △BDC ~△ABC

Therefore,

CD BC

----- = -----

BC AC

(Condition for similarity)

Or, BC²= CD × AC …………………………..(2)

Adding the equations (1) and (2) we get,

AB² + BC² = AD × AC + CD × AC

AB² + BC² = AC (AD + CD)

Since, AD + CD = AC

Therefore, AC² = AB² + BC²

Hence, the Pythagorean thoerem is proved.

and

Construction- Extend the line segment DE and produce it to F such that, EF=DE.

In the triangle, ADE, and also the triangle CFE

EC= AE —– (given)

∠CEF = ∠AED {vertically opposite angles}

EF = DE { by construction}

hence,

△ CFE ≅ △ ADE {by SAS}

Therefore,

∠CFE = ∠ADE {by c.p.c.t.}

∠FCE= ∠DAE {by c.p.c.t.}

and CF = AD {by c.p.c.t.}

The angles, ∠CFE and ∠ADE are the alternate interior angles. Assume CF and AB as two lines which are intersected by the transversal DF.

In a similar way, ∠FCE and ∠DAE are the alternate interior angles. Assume CF and AB are the two lines which are intersected by the transversal AC.

Therefore, CF ∥ AB

So, CF ∥ BD

and CF = BD {since BD = AD, it is proved that CF = AD}

Thus, BDFC forms a parallelogram.

By the use of properties of a parallelogram, we can write

BC ∥ DF

and BC = DF

BC ∥ DE

and DE = (1/2 * BC).

Hence, the midpoint theorem is Proved.