Question

Pythagoras theorem and mid point theorem​

Answer 1

ʜᴇyᴀ ᴅᴜᴅᴇ..✌

ᴜʀ ᴀɴꜱᴡᴇʀ ↓

Mid-point theorem

Prove that the line joining the mid-points of two sides of a triangle is parallel to the third side and equal to half the length of the third side.

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ᴩʟᴢ ᴍᴀʀᴋ ᴀꜱ ʙʀᴀɪɴʟɪᴇꜱᴛ☃

Answer 2

$\huge\underline\mathfrak\red{Pythagoras \: therom}$

• 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 triangle 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 a, b and c) which have positive integer values, when squared, are put into an equation, also called a Pythagorean triple.

$\huge\underline\mathcal\purple{formula}$

Consider the triangle given above:

Where “a” is the perpendicular side,

“b” is the base,

“c” is the hypotenuse side.

According to the definition, the Pythagoras Theorem formula is given as:

${hypotenuse}^{2} = {base}^{2} + {perpendicular}^{2} \\ {c}^{2} = {a}^{2} + {b}^{2}$

$\huge\underline\mathfrak\red{Mid \: point \: therom }$

• If the line segment adjoins midpoints of any of the sides of a triangle, then the line segment is said to be parallel to all the remaining sides, and it measures about half of the remaining sides.

$\huge\underline\mathcal\purple{Proof}$

Consider the triangle ABC, as shown in the above figure,

Let E and D be the midpoints of the sides AC and AB. Then the line DE is said to be parallel to the side BC, whereas the side DE is half of the side BC; i.e.

DE∥BC

DE = (1/2 * BC).

Now consider the below figure,

Mid- Point Theorem

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

In triangle ADE and CFE,

EC = AE —– (given)

∠CEF = ∠AED (vertically opposite angles)

EF = DE (by construction)

By SAS congruence criterion,

△ CFE ≅ △ ADE

Therefore,

∠CFE = ∠ADE {by c.p.c.t.}

∠FCE= ∠DAE {by c.p.c.t.}

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

∠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 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.

${\huge{\fcolorbox{black}{WHITE}{♡Itz \: AISH♡}}}$