**Bad school buildings**

Mrs. Sanyal was the founding principal of the only school in the slum of Jagatbasti. Yes, it was a big achievement that the school had survived here for over a decade despite facing every hurdle. As a principal, Mrs. Sanyal was proud that the school had made it this long under her leadership. She was not happy about that the school buildings were in very poor condition. Roofs of several of the rooms leaked during the rainy season. That was risky and demoralising for the students. She had approached the authorities many a time but it seemed that they would rather see the school closed even though there was no other school nearby.

**A kind but careful donor**

In an attempt to save the school, Mrs. Sanyal had a meeting with a young industrialist named Mr, Avish Dodi. Mr. Dodi had just started a large factory on the outskirts of the town. He wanted to have good public image of himself as an industrialist. He had come to see the school at Mrs. Sanyal’s request. He agreed to donate Rs. 100,000. Mrs. Sanyal did not seem happy as this amount would not solve the problem of repairing leaky roofs of the school. It would require a much larger amount.

Mr. Dodi: Look Mrs. Sanyal. Let me be straight. I am not too thrilled about the school training in which kids are trained to memorize everything and regurgitate it. I am still willing to contribute Rs. 100,000.

**Ms. Rania Ali**

Mrs. Sanyal took a deep breath and asked the peon to get Ms. Rania Ali. The young enthusiastic teacher Rania Ali came. Mrs. Ali knew that the kids had, at best, only marginal interest in learning, She was aware of her challenge but the great teacher found new ways to get attention of her students. She found pride in connecting with the students and teaching them. The principal was aware of her superb qualities.

Mrs. Sanyal: Rania, Mr. Dodi thinks that we are training parrots here. The students have no understanding of what we teach them. You have been very innovative in your teaching geometry. Can you please tell him so ?

Mr, Dodi: You have probably told them to memorize proof of every theorem. That does not mean that they understood anything.

Mrs. Rania Ali paused, smiled and then said: The students of my geometry class will prove you wrong but it will cost you.

Mr. Dodi: What do you mean?

Mrs. Rania Ali: You must have learned about similar and congruent triangles but I don’t know how much. My students learned about similar triangles lst week. What will you give if they can prove the Pythagoras theorem using this concept ?

Mr. Dodi: What would they do different from what is in their textbook ?

Mrs. Rania Ali: If you can spend the day with us tomorrow, student after student can come and give you a different proof of the theorem. No proofs will be repeated but it will cost you. For one proof you may stick to a donation of Rs. 100,000 but double the donation after each additional proof they come up with. I said, it will cost you.

Mr. Dodi: I like that. If they can do that, I will keep doubling the amount of donation to a maximum of 100 times the original amount. See you at 9 am in your class tomorrow.

Mrs. Rania Ali: Please, bring your cheque book. Excuse me, now I have one hour to go prepare the students to rob you tomorrow.

Mrs. Rania Ali asked all her geometry class students to meet. She discussed the issue with them. She told them that if they succeeded there would be a big party afterwards. Taheen was excited. He loved the idea and immediately came up with several constructions. Mrs. Rania Ali discussed with the students on how to approach the problem of coming up with the proofs. They worked in groups. Of course, everyone was ready to work for the big party.

**Next morning**

Next morning, Mr. Dodi came to the school with a big smug on his face. He had learnt only one proof fot this theorem. He thought the students could not do too much more than that. The first student was Mahek and she gave the proof that Mr. Dodi had learned in his school. So the man smiled. By lunch time, however, Rehan, Rajab and Arisha had each presented a new proof that he had never seen before. He was impressed. All the kids did was to rotate and rearrange similar or congruent right angle triangles to make different shapes. Then they used these constructions to prove the theorem. There were four different proofs. He would have to donate Rs. 800,000 since the amount doubled with each student. After lunch came Fateh, Feran, Anya and Taheen. That was a total of 8 students each with a new proof for the theorem. Had the amount doubled 5 times, he would have to give 256 times the original amount. As he had set the limit of 100 times, the original amount of Rs, 100,000. Mr. Dodi gave a cheque for Rs. 10,000,000 to the school.

Not only the roofs were repaired, there was money left for a big party and prizes for the students who had contributed.

Challenge: The eight proofs that the class gave follow. Please, go over them with friends in your class. Now show me that you are as smarter than this class. Find a ninth proof of the theorem on your own or by chatting with your friends. Apparently, there are over 180 proofs on the internet but the challenge is to find one on your own. Who knows, you may invent a new one. Show the proof to your parents with pride. Best of luck.

**THE EIGHT PROOFS SHOWN TO MR. DODI. **

Pythagoras theorem states that for a right ∠ Δ the hypotenuse square equals base square plus height square.

**1. Mehak gave the proof that they had been taught – the one given by the 20th US President James Garfield in 1876..**

**Construction: ** ABC is a right ∠ Δ with the base AB = a, height AC = b and the hypotenuse BC = c.

Extend the line AC to D so that the line segment CD is of length a (same as the base). Draw a right ∠ CDE from CD so that DE = b.

Because CDE and ABC are congruent Δs, CE = c and ∠ DCE = ∠ ABC. ∠ BCE = 180° – ∠ ACB – ∠ DCE (same as ABC) = 90°.

Therefore, Δ BCE is a right ∠ Δ.

**Proof:**

Area of the trapezoid ABDE = height x (base + top)/2 = (a+b)(a+b)/2.

The area of the trapezoid ABDE is the sum of the areas of the right ∠ Δs ABC + CDE +BCE.

Therefore, the trapezoid area = ab/2 + ab/2 + c^{2}/2 = ab + c^{2}/2.

Therefore, (a+b)(a+b)/2 = ab + c^{2}/2 or (a+b)(a+b)

= 2 ab + c^{2} or a^{2} +b^{2}+2ab

=2 ab + c^{2} or a^{2} + b^{2 }= c^{2}.

This proves the Pythagoras theorem that for a right ∠ Δ,

base^{2 }+ height^{2 }= hyotenuse^{2}.

**2. , Rehan rotated four congruent right ∠ Δs to make an inner and an outer square**

**Construction: ** Take a right ∠ Δ with the sides a and b and the hypotenuse c.

Rotate the Δ clock wise by 90⁰ to get a second Δ.

Rotate it again to get the third and then again to get the fourth.

Thus you have rotated the original Δ by 90, 180 and 270⁰.

Put the 4 Δs together so that they form a square of the length c and an internal square with the length a-b.

**Proof: ** Area of the large square of side c = c^{2} and that of the internal square of side a-b = (a-b)^{2}

Area of each Δ = ab/2 and the area of the four Δs = 2ab.

The area of the large rectangle is the sum of areas of the four Δs (2ab) and the inner square (b^{2}).

That means c^{2 }= (a-b)^{2} + 2ab or c^{2 }= a^{2}+b^{2}-2ab + 2ab

or c^{2 }= a^{2}+b^{2} as stated by Pythagoras theorem.

**3. Rajab also rotated 4 identical right ∠ Δs but arranged them differently**

**Construction: ** Rajab drew a right ∠ Δ with base a, height b and hypotenuse c.

Then he made three congruent Δs by rotating the original Δ by 90⁰, 180⁰ and 270⁰.

He placed the triangles together so that they made an outer square of side a + b and an internal square of sides c.

**Proof: ** The area of the outer square = (a+b)^{2}.

The area of the outer square can also be written as the sum of areas of the inner square (c^{2}) and the four Δs ( 4 x ab/2 oe 2ab).

Then (a+b)^{2} = c^{2} + 2ab

or a^{2 +} b^{2} + 2ab = c^{2} + 2ab

or a^{2 }+ b^{2} = c^{2} as stated by the Pythagoras theorem.

**4. Arisha just drew a right ∠ Δ and a vertical on its hypotenuse**

**Construction:** ABC is a right ∠ Δ with ∠ CAB = 90⁰.

### Draw a vertical AD from point A to its hypotenuse BC.

**Proof:** In Δ ABC, ∠ CAB = 90⁰ and the sum of ∠s ACB and CBA = 180⁰ – 90⁰ = 90⁰.

### For Δ ADC, ∠ ADC = 90⁰, ∠ ACB is common with the Δ CAB.

### Therefore, ∠ CAD = ∠ ABD.

### Thus Δs CAB and ADC are similar.

### Similarly, Δ ABD is also similar to these two Δs,

### Therefore AB/BC = BD/AB……………. eq. 1

### and AC/BC = DC/AC………. eq 2.

### Rewriting eq 1 AB/BC = BD/AB

### Muliplying both sides by AB.BC

### AB^{2 }= BC.BD ………………eq 3

### Similarly from eq 2 AC/BC = DC/AC, we get AC^{2} = BC.DC ………….. eq 4

### Summing up equations 3 and 4 gives

### AB^{2 }+ AC^{2} = BC.BD + BC.DC or

### AB^{2 }+ AC^{2} = BC (BD +DC)

### Since BD + DC = BC (see figure)

### AB^{2 }+ AC^{2} = BC. BC or

### AB^{2 }+ AC^{2} = BC^{2} (Pythagoras theorem).

**5. Fateh drew a main right ∠ Δ and a similar Δ to its left side **

**Construction:** Draw a Δ ABC with the ∠ CAB being 90⁰.

### Draw a Δ ABD similar to ABC such that line CA continues to D, ∠ DAB = 90⁰, ∠ ADB = ∠ ABC and ∠ ABD = ∠ ACB.

### Then the larger Δ DBC is also similar to Δ ABC,

### Proof: For similar Δs

### AD/AB = AB/AC

### AD = AB^{2}/AC

### BD/AB = BC/AC

### BD = AB·BC/AC

### Areas of Δ DAB + Δ ABC = Δ DBC

### Area of Δ ABD = AB.AD/2 = (AB²/AC)·AB/2

### Area of Δ ABC = AB.AC/2

### Area of Δ BCD = (AD+AC).AB/2

### Therefore, AD.AB + AC.AB = BD. BC

### AD.AB + AC.AB = (AB·BC/AC)·BC

### AB. AB^{2}/AC + AC.AB = (AB·BC/AC)·BC

### (AB·BC/AC)·BC = AB. AB^{2}/AC + AC.AB

### BC². AB/AC = AB. AB^{2}/AC + AC.AB Divide by AB/AC

### BC² = AB^{2} + AC^{2 }(Pythagoras theorem)

### 6. Feran’s similar right ∠ Δs had tricky lengths

**Construction: ** Make a right ∠ Δ ABC with the hypotenuse AC of length cc (c . c) and the sides BC of length ac and AB of length bc.

### Construct another right ∠ Δ ADB with the hypotenuse AB of length bc and lengths side AD of length ab and DB of length bb,

### and another right ∠ Δ BCE of the hypotenuse BC of length cc and sides BE and CE lengths aa and ab.

### Lay them together so as to form a rectangle ADEC as shown.

**Proof: **The rectangle ADEC with the side AC of length cc (c^{2}) and side DE = aa + bb (a^{2}+b^{2}).

### Opposite sides of a rectangle are of equal length.

### Therefore c^{2} = a^{2 }+ b^{2} (Pythagoras theorem).

### 7**. Anya drew a right ∠ Δ ABC and a congruent right ∠ Δ BDF side ways so that BD layed on AB**

**Construction:** Let ABC and BED be identical right ∠ Δs, with point E lying on AB. Join points A and D to form the Δ ABD.

### Proof: Evaluate the area of ΔABD in two ways:

Area(ΔABD) = BD·AF/2 = AB.DE /2.

Using the notations as indicated in the diagram:

c(c – x)/2 = b·b/2.

x = CF can be found by noting the similarity (BD is perpendicular to AC) of Δs BFC and ABC:

x = a²/c.

c(c-a^{2}/c)/2 = b^{2}/2 or

c(c-a^{2}/c) = b^{2} or

c^{2 }– a^{2 }= b^{2}

or c^{2 }= a^{2} + b^{2} (Pythagoras theorem)

** 8. Taheen’s proof involved some algebra**

**Construction: **Let ABC and DEF be two congruent right ∠ Δs such that B lies on DE and A, F, C, E are collinear.

Let BC = EF = a,

AC = DF = b,

The hypotenuses AB = DE = c.

Obviously, AB is perpendicular to DE because ABE is a right ∠.

**Proof: **Join the points A and D. Compute the area of ΔADE in two different ways.

*In the first step, determine the length of CE.*

CE can be found from similar Δs BCE and DFE:

CE/FE = BC/DF

or CE = BC·FE/DF = a·a/b = a²/b.

*In the second steo, compute the area of ΔADE in two different ways.*

Area (ΔADE) on base DE = AB·DE/2 = c²/2………………eq1

And Area(ΔADE) on base AE = DF·AE/2 = b·AE/2 = b,

or (AC +CE)/2 = b.(b+ a²/b)/2………………. eq. 2

Now the two areas of ΔADE must be the same. Therefore,

c²/2 = b(b + a²/b)/2

or c² = b(b + a²/b)

or c² = b^{2} + a^{2}, thus proving the Pythagorus theorem.