Welcome to today's mathematics lesson. We are going to explore one of the most beautiful and powerful theorems in geometry — the Pythagoras Theorem. This theorem connects the three sides of a right-angled triangle in a remarkable way, and we will see not just why it works, but also how to use it and how to work backwards from it.
Let us begin with the theorem itself. In a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.
Here is what this means. Imagine a right-angled triangle ABC, where the angle at B is 90°. The side opposite this right angle — AC — is called the hypotenuse, and it is always the longest side. The theorem tells us that if we construct squares on all three sides, then the area of the square on the hypotenuse equals the combined areas of the squares on the other two sides. Algebraically, this becomes: AC² = AB² + BC².
Now, let me walk you through a beautiful area-based proof of this theorem. We start with our right-angled triangle ABC, with the right angle at B. On each side, we construct a square: ACDE on the hypotenuse, ABFG on side AB, and BCHI on side BC.
Next, we draw a line from B perpendicular to the hypotenuse AC, extending it to meet the opposite side of the large square. This creates two rectangles inside the square on the hypotenuse.
Here is the clever observation. Consider triangles GAC and BAE. Angle GAC equals angle BAE — both equal 90° + x, where x is angle BAC. Also, AG = AB and AC = AE, being sides of squares. Therefore, these triangles are congruent by the side-angle-side criterion.
Now, triangle GAC and the square ABFG share the same base AG and lie between the same parallels. So the area of triangle GAC is exactly half the area of square ABFG. Similarly, triangle BAE and the first rectangle share base AE and the same parallels, so triangle BAE has half the area of that rectangle.
Since the triangles are congruent, their areas are equal. Therefore, the square on AB equals the first rectangle. By an identical argument, the square on BC equals the second rectangle. Adding these, the two squares on the shorter sides together equal the square on the hypotenuse. Hence, AB² + BC² = AC².
There is also an elegant alternative proof using similar triangles. From the right angle at B, we drop a perpendicular to the hypotenuse AC, meeting it at point D.
This creates two smaller triangles, each similar to the original triangle and to each other. From the similarity of triangles ABC and BDC, we get BC² = AC × DC. From the similarity of triangles ABC and ADB, we obtain AB² = AC × AD.
Adding these two equations: AB² + BC² = AC × AD + AC × DC. Factoring out AC, this becomes AC × (AD + DC), which is simply AC × AC = AC². Thus we have proved the theorem again, this time through the powerful lens of similarity.
Now, what about the converse? The converse is equally important and gives us a test for right angles. If in any triangle, the square on the largest side equals the sum of the squares on the other two sides, then the triangle must be right-angled, with the right angle opposite the largest side.
More generally, this relationship tells us about the type of triangle. If the square of the largest side equals the sum of the other two squares, we have a right-angled triangle. If it exceeds the sum, the triangle is obtuse-angled. If it is less than the sum, the triangle is acute-angled.
Let us apply this to a practical problem. A ladder 17 m long reaches a window 15 m above the ground. When pivoted at the same base point, it reaches 8 m high on the opposite side of the street. We need to find the street's width.
First position: the ladder forms a right triangle with height 15 m and hypotenuse 17 m. By the theorem, the horizontal distance is √(17² − 15²), which equals √(289 − 225), giving √64 or 8 m.
Second position: height is 8 m, hypotenuse still 17 m. The horizontal distance becomes √(17² − 8²), which is √(289 − 64), giving √225 or 15 m.
The total street width is the sum: 8 m + 15 m = 23 m. Notice the beautiful symmetry — the two horizontal distances swap the values of the two heights.
Here is another elegant application. We are given that AB = 18 cm and CD = 6 cm, with BP = 12 cm and CP = 9 cm. We must prove that angle APD is 90°.
We calculate AP² from triangle APB: 18² + 12² = 324 + 144 = 468. We calculate DP² from triangle DPC: 6² + 9² = 36 + 81 = 117. Their sum is 585.
Now we find AD² by constructing a perpendicular from D to AB. The vertical separation is 21 cm and the horizontal is 12 cm. Thus AD² = 12² + 21² = 144 + 441 = 585.
Since AP² + DP² = AD², the converse of Pythagoras Theorem confirms that angle APD is indeed 90°.
Let me introduce you to Pythagorean triplets. These are sets of three positive integers that satisfy the theorem. The most famous is 3, 4, 5, since 3² + 4² = 9 + 16 = 25 = 5². Any multiple of these also works — so 6, 8, 10 or 9, 12, 15 are also triplets. Other common triplets include 5, 12, 13 and 8, 15, 17. Recognizing these can save you calculation time in problems.
Now let us explore some important extensions and proofs involving the theorem.
First, consider a triangle where AD is perpendicular to BC produced. We can prove that c² = a² + b² + 2ax, where x is the extension beyond C. Using the theorem on triangle ABD: c² = (a + x)² + h². Expanding and substituting x² + h² = b² from triangle ADC, we obtain the result. This is useful when dealing with obtuse-angled configurations.
Next, consider points M and N on the sides of a right-angled triangle PQR, right-angled at Q. We can prove that PM² + RN² = PR² + MN². The key is to apply the theorem to triangles PQM and NQR, then rearrange using the fact that PR² = PQ² + QR² and MN² = QM² + NQ².
Here is a beautiful result about the altitude to the hypotenuse. If CD = p is perpendicular to hypotenuse AB in a right-angled triangle, then 1/p² = 1/a² + 1/b². We prove this by noting that the area equals both ½ab and ½cp, so c = ab/p. Substituting into c² = a² + b² and dividing by a²b² gives the elegant reciprocal relationship.
Let us prove a result about an equilateral triangle. If ABC is equilateral and P divides BC in ratio 2:1, then 9AP² = 7AB².
Draw the altitude AD. Then BD = ½BC = ½AB, and BP = ⅔AB. So DP = BP − BD = ⅔AB − ½AB = (1/6)AB.
From triangle ABD: AD² = AB² − AB²/4 = (3/4)AB². From triangle ADP: AP² = AD² + DP² = (3/4)AB² + AB²/36. This simplifies to (28/36)AB² = (7/9)AB², giving 9AP² = 7AB².
Finally, one of the most elegant results in quadrilateral geometry. In any parallelogram, the sum of the squares of the diagonals equals the sum of the squares of all four sides.
For parallelogram ABCD, we prove AC² + BD² = AB² + BC² + CD² + DA². Draw perpendiculars from A and D to appropriate sides. Apply the theorem systematically to right triangles formed, and use the properties that opposite sides of a parallelogram are equal.
Special cases follow immediately. For a rhombus, all sides are equal, so 4 × (side)² = sum of squares of diagonals. For a rectangle, the diagonals are equal, so length² + breadth² = diagonal², which is just the Pythagoras theorem itself.
Let us recap the key takeaways from today's lesson.
First, the Pythagoras Theorem states that in a right-angled triangle, the square on the hypotenuse equals the sum of the squares on the other two sides: c² = a² + b².
Second, the converse is equally powerful — if the square on one side equals the sum of squares on the other two, that angle is a right angle.
Third, this relationship classifies triangles: equal means right, greater means obtuse, less means acute.
Fourth, Pythagorean triplets like 3, 4, 5 are integer solutions that appear frequently in problems.
Fifth, the theorem extends beautifully to prove results about altitudes, medians, and points inside triangles.
Sixth, in any parallelogram, the sum of squares of diagonals equals the sum of squares of all sides, unifying several geometric properties.
Mathematics is built on such elegant connections. The Pythagoras Theorem, discovered independently across ancient civilizations, remains one of the most useful tools in geometry. Practice applying it to varied situations, and you will find right triangles hidden everywhere. Until next time, keep exploring, keep proving, and enjoy the beauty of mathematical reasoning.