Theorems Archives – Mathemafia

State and Prove Pythagoras Theorem

Sunil Kumar — Tue, 11 Mar 2025 11:29:49 +0000

Pythagoras Theorem

In a right angled triangle, the square of hypotenuse is equal to the sum of squares of the other two sides.

In a right angled triangle, the square of hypotenuse is equal to the sum of squares of perpendicular and base of the triangle.

To Prove

$ \begin{aligned} &BC^2 =AB^2+AC^2\end{aligned}$

Construction

$ \begin{aligned} &\text{Draw } AD \perp BC\end{aligned}$

Proof

Part 1

$ \begin{aligned} &In \ \ \triangle ABC \text{ and } \triangle ADB \end{aligned}$

$ \begin{aligned} &\angle {BAC} = \angle {ADB} = 90^{\circ}\\ &\angle B = \angle B \ \ \ \ \text{(Common)} \end{aligned}$

By AA similarity criterion

$ \begin{aligned} &\triangle ABC \sim \triangle DBA \end{aligned}$

Very Important: We write it as △ABC is similar to △DBA. We cannot write it similar to △ADB even though △DBA and △ADB are same triangles.

Why? Click here to read it or watch the video embedded below.

$ \begin{aligned} &\Rightarrow \frac{AB}{DB} = \frac{BC}{AB}\\\\ &\Rightarrow AB^2 = BC \times DB \ \ …(1)\end{aligned}$

Part 2

$ \begin{aligned} &In \ \ \triangle ABC \text{ and } \triangle ADC \end{aligned}$

$ \begin{aligned} &\angle {BAC} = \angle {ADC} = 90^{\circ}\\ &\angle C = \angle C \ \ \ \ \text{(Common)} \end{aligned}$

By AA similarity criterion

$ \begin{aligned} &\triangle ABC \sim \triangle DAC \end{aligned}$

$ \begin{aligned} &\Rightarrow \frac{BC}{AC} = \frac{AC}{DC}\\\\ &\Rightarrow AC^2 = BC \times DC \ \ …(2)\end{aligned}$

Part 3

Adding (1) and (2)

$ \begin{aligned} AB^2+ AC^2 &= BC \times BD + BC \times DC \\ &=BC \times (BD + DC)\\ &=BC \times BC\\ &=BC^2\\ \end{aligned}$

Hence proved.

Watch on YouTube

Why?

$ \begin{aligned} &\text{Why }\triangle ABC \sim \triangle DBA \text{ but } \triangle ABC \not\sim \triangle ADB\end{aligned}$

When it comes to similarity and congruence of triangles, students always find it slightly confusing but let me tell you, it is as simple as Virat Kohli hits a six 🏏 at will. All you need is to possess good mathematics skills 😜 like Kohli has (in cricket of-course).

Ok … That’s not true. You don’t need to be as good as Kohli 😜 to understand this concept. 💃🕺 Wow!!! Nacho Nacho. 💃🕺 (If this song can win oscar, one day I will get the Fields Medal. 🏅 🏆 )

Explanation

We just proved that the triangles shown below are similar. The triangles are similar when all three corresponding angles are congruent in both the triangles. In this case:

$ \begin{aligned} \angle {BAC} &= \angle {ADB} = 90^{\circ}\\ \angle B &= \angle B \ \ \ \ \text{(Common)}\\ \Rightarrow \angle {ACB} &= \angle {DAB} \end{aligned}$

In the picture displayed above you can clearly see mapping of the corresponding angles means the mapping of the corresponding vertices of the two triangles. So while writing the similarity expression, we must write the corresponding vertices in same order on both sides as shown in Mapping Pic.

Welldone Mathemafia 👏💯 !!!

Pytha: Oh my Goras!!! I can’t believe that you have explained it so Beautifully Colorfully.

Mathemafia: Wow! You struck out Beautifully. 💔

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SSA is not a valid congruence criterion

Sunil Kumar — Sun, 09 Mar 2025 12:17:43 +0000

SSA is not a valid congruence criterion

Student: What is SSA?

Mathemafia – I call it Side Side Aaaaaaaaaa…..hhhhhh !!!

Not funny, huh!! But effective 🎯 enough to remember 🧠.

If you want to read, it may take 5 minutes. If you want to save time, watch the short video and then send me half 🌓 of the money you save.

You know… ⏰ Time is 💰 Money 😂 🤣 😆

If you still insist, written below is the explanation why SSA is not a valid congruence criterion. 🤿 👀 ❤️ 💃

But First, Let’s Sing it!!!

As I always say – Mathematics is like Music. Enjoy it!!! But don’t enjoy this one alone 🥳. This piece of music (SSA) is a party 🎉 song 🎶. So enjoy it with your friends. 💃🕺💃🕺. (I can’t ask in more creative 🎭 way to share 🔗 it. Hmmmmm… Let me say it clearly then… I BEG YOU. SHARE it pls. 😭 🙏 😭🙏 )

SSA – Sing Sing Aaaaaaaa…..

SSA – Share Share Aaaaaaaa…..

SSA – Subscribe Subscribe Aaaaaaaa…..

Student – Sir, how about if we reverse the abbreviation 🤭 🙈 😆 🍑 🫏

Mathemafia – Wow!!! That’s much more creative. 🫏 🫏

Explanation

Let us draw an obtuse angled triangle, ABC with obtuse angle at vertex A as shown in the Pic 1.

Obviously side BC will be the longest side.

Side opposite to largest angle is the longest side in a triangle.

Take a compass and draw an arc of length AB. It will intersect the longest side BC, say at point D.

See Pic 2.

Join AD.

Let us now focus on two triangles▲ABC and ▲ADC

$ \begin{aligned} \overline{AB} &\cong \overline{AD} \ \ (Construction) \\ \overline{AC} &\cong \overline{AC} \ \ (Common) \\ \angle C &= \angle C \ \ (Common) \end{aligned}$

So we got two congruent sides and one congruent angle in SSA formation but clearly the two triangles are not congruent. Hence, SSA is not a valid congruent criterion.

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Proof | Integers are infinitely many

Sunil Syal — Tue, 22 Sep 2020 13:42:31 +0000

Proof - Integers are Infinitely Many

Theorem: Prove that there are infinitely many integers.

Proof: Let us assume that there are finitely many Integers.

$ \begin{aligned} &\Rightarrow \text{∃ a greatest integer, say } x \end{aligned}$

Let us get a new number say y by adding 1 to x.

$ \begin{aligned} y&=x+1\end{aligned}$

$ \begin{aligned} &Now \ \ \mathbf{y} \in\mathbb{Z} \\ &\text{because } \mathbf{x+1} \text{ is also an integer}\\\\&But \ \ y > x \end{aligned}$

So we got another integer y > x.

This is a contradiction to our assumption that x is the greatest integer.

This means our assumption is wrong that there are finitely many integers.

Hence we can conclude that there are infinitely many integers.

Watch on YouTube

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State and Prove Remainder Theorem – Polynomials

Sunil Syal — Tue, 15 Sep 2020 06:39:56 +0000

Remainder Theorem

Let p(x) be any polynomial of degree greater than or equal to one and let a be any real number. If p(x) is divided by the linear polynomial x – a, then the remainder is p(a).

Proof

Let p(x) be any polynomial of degree >= 1

Dividing p(x) by x-a we will get

$ \begin{aligned} &p(x) = (x-a)q(x)+ r(x) \\\\ &\text{or we can write}\\\\&p(x) = (x-a)q(x)+ r \ \ …(1)\end{aligned}$

Where q(x) and r are quotient and remainder respectively.

Remainder r will be a constant because either r is 0 or deg(r) < deg (x-1).

Not able to get it? Watch the video tutorial or Click here to read Why Remainder is a constant?

Put x=a in equation (1)

$ \begin{aligned} p(a) &= (a-a)q(a)+ r \\ \Rightarrow \ \ p(a) &= 0+ r \\ \Rightarrow \ \ r &= p(a) \end{aligned}$

Hence proved.

Watch on YouTube

Why Remainder is a constant?

The deg (x-a) is 1. If we divide any polynomial of deg >= 1 by x – a, there will be two possibilities:

It will be divided completely and the remainder will be 0
Remainder will be non-zero

If the remainder is 0, then we are happy 💃🕺 with what we want to prove. Obviously, 0 is also a constant.

Let us now talk 📣 🗣️ about the second scenario – Remainder is non-zero.

We can keep on dividing a polynomial with x – a until we get a remainder that has degree smaller than degree of x – a.

Only when deg (r) < deg (x – a), we can’t divide further.

This means the deg (r) = 0 because deg (x – a) = 1. This means remainder is a constant.

If this is still not clear then either I am not able to explain it well 😵‍💫 🥀 … or you … 😜

More confusion 🤷‍♂️ 🤷‍♀️ Does this mean the Deg (r) = 0?

As we know, that remainder can be

(1) Non-Zero Constant

(2) Zero

(1) Non-Zero Constant

The degree of any non-zero constant is 0. You can understand it like this:

$ \begin{aligned} &3 \text { can be written as } 3x^0 \end{aligned}$

(2) Zero

The degree of 0 is not defined. You can understand it like this:

$ \begin{aligned} &0 \text { can be written as } 0x^0 = 0x^3=0x^7=0x^{565} \end{aligned}$

Final Deal 💰

After explaining so much if this is still not clear then you need to either say me some “good words” 🤬 $#%&%#🚫 or you must invite me to a have a cup of coffee ☕ so that I can explain it better 👨‍🏫 / 👩‍🏫. Mind it – I will not go out of Gurgaon 🚗 for coffee. 😎

Prerequisite

This is a very important theorem from polynomials perspective. It helps solve very long divisions quite quickly. To understand its significance, you are recommended to learn about the long division of polynomials. That will give you more clarity why this theorem is so important and helpful.

(Topic 2.4 of chapter 2 of class 9 Mathematics NCERT textbook)

Long division method

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