prove that in a right angle triangle the square of the hypotenuse is equal to the sum of the squares on the other two sides
Answers
Answered by
3
Proof using similar triangles

Proof using similar triangles
This proof is based on the proportionality of the sides of two similar triangles, that is, upon the fact that the ratio of any two corresponding sides of similar triangles is the same regardless of the size of the triangles.
Let ABC represent a right triangle, with the right angle located at C, as shown on the figure. Draw the altitude from point C, and call H its intersection with the side AB. Point H divides the length of the hypotenuse c into parts d and e. The new triangle ACH is similar to triangle ABC, because they both have a right angle (by definition of the altitude), and they share the angle at A, meaning that the third angle will be the same in both triangles as well, marked as θ in the figure. By a similar reasoning, the triangle CBH is also similar to ABC. The proof of similarity of the triangles requires the triangle postulate: the sum of the angles in a triangle is two right angles, and is equivalent to the parallel postulate. Similarity of the triangles leads to the equality of ratios of corresponding sides:
{\displaystyle {\frac {BC}{AB}}={\frac {BH}{BC}}{\text{ and }}{\frac {AC}{AB}}={\frac {AH}{AC}}.}
The first result equates the cosines of the angles θ, whereas the second result equates their sines.
These ratios can be written as
{\displaystyle BC^{2}=AB\times BH{\text{ and }}AC^{2}=AB\times AH.}
Summing these two equalities results in
{\displaystyle BC^{2}+AC^{2}=AB\times BH+AB\times AH=AB\times (AH+BH)=AB^{2},}
which, after simplification, expresses the Pythagorean theorem:
{\displaystyle BC^{2}+AC^{2}=AB^{2}\ .}
The role of this proof in history is the subject of much speculation. The underlying question is why Euclid did not use this proof, but invented another. One conjecture is that the proof by similar triangles involved a theory of proportions, a topic not discussed until later in the Elements, and that the theory of proportions needed further development at that time.[12][13]
hope it helps u

Proof using similar triangles
This proof is based on the proportionality of the sides of two similar triangles, that is, upon the fact that the ratio of any two corresponding sides of similar triangles is the same regardless of the size of the triangles.
Let ABC represent a right triangle, with the right angle located at C, as shown on the figure. Draw the altitude from point C, and call H its intersection with the side AB. Point H divides the length of the hypotenuse c into parts d and e. The new triangle ACH is similar to triangle ABC, because they both have a right angle (by definition of the altitude), and they share the angle at A, meaning that the third angle will be the same in both triangles as well, marked as θ in the figure. By a similar reasoning, the triangle CBH is also similar to ABC. The proof of similarity of the triangles requires the triangle postulate: the sum of the angles in a triangle is two right angles, and is equivalent to the parallel postulate. Similarity of the triangles leads to the equality of ratios of corresponding sides:
{\displaystyle {\frac {BC}{AB}}={\frac {BH}{BC}}{\text{ and }}{\frac {AC}{AB}}={\frac {AH}{AC}}.}
The first result equates the cosines of the angles θ, whereas the second result equates their sines.
These ratios can be written as
{\displaystyle BC^{2}=AB\times BH{\text{ and }}AC^{2}=AB\times AH.}
Summing these two equalities results in
{\displaystyle BC^{2}+AC^{2}=AB\times BH+AB\times AH=AB\times (AH+BH)=AB^{2},}
which, after simplification, expresses the Pythagorean theorem:
{\displaystyle BC^{2}+AC^{2}=AB^{2}\ .}
The role of this proof in history is the subject of much speculation. The underlying question is why Euclid did not use this proof, but invented another. One conjecture is that the proof by similar triangles involved a theory of proportions, a topic not discussed until later in the Elements, and that the theory of proportions needed further development at that time.[12][13]
hope it helps u
Answered by
1
Answer:
Given :
A right triangle ABC right angled at B.
To prove :
AC² = AB² + BC²
Construction :
Draw BD ⊥ AC
Proof :
In Δ ADB and Δ ABC
∠ A = ∠ A [ Common angle ]
∠ ADB = ∠ ABC [ Both are 90° ]
∴ Δ ADB Similar to Δ ABC [ By AA similarity ]
So , AD / AB = AB / AC [ Sides are proportional ]
= > AB² = AD . AC ... ( i )
Now in Δ BDC and Δ ABC
∠ C = ∠ C [ Common angle ]
∠ BDC = ∠ ABC [ Both are 90° ]
∴ Δ BDC Similar to Δ ABC [ By AA similarity ]
So , CD / BC = BC / AC
= > BC² = CD . AC ... ( ii )
Now adding both equation :
AB² + BC² = CD . AC + AD . AC
AB² + BC² = AC ( CD + AD )
AB² + BC² = AC² .
AC² = AB² + BC² .
Hence proved .
Attachments:
Similar questions