File:Pythagoras similar triangles simplified.svg
Summary
Simplified version of <a href="https://en.wikipedia.org/wiki/Pythagorean_theorem#Proof_using_similar_triangles" class="extiw" title="w:Pythagorean theorem">similar triangles proof</a> for <a href="//commons.wikimedia.org/wiki/Category:Pythagorean_theorem" title="Category:Pythagorean theorem">Pythagoras' theorem.</a>
In triangle ACB, angle ACB is the right angle. CH is a perpendicular on hypotenuse AB of triangle ACB.
In triangle AHC and triangle ACB, ∠AHC=∠ACB as each is a right angle. ∠HAC=∠CAB as they are common angles at vertex A. Thus triangle AHC is similar to triangle ACB by <a href="https://en.wikipedia.org/wiki/Similarity_(geometry)#Angle.2Fside_similarities_for_triangles" class="extiw" title="w:Similarity (geometry)">AA test.</a>
Thus, https://wikimedia.org/api/rest_v1/media/math/render/svg/edcf1ae6eaca6325cc257f26561db4c9aa6c499a" class="mwe-math-fallback-image-inline mw-math-element" aria-hidden="true" style="vertical-align: -2.005ex; width:12.139ex; height:5.509ex;" alt="{\displaystyle {\frac {AC}{AB}}={\frac {AH}{AC}}}"> <img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/89edfb090cfb5ac5f9e4465566602cb0526af347" class="mwe-math-fallback-image-inline mw-math-element" aria-hidden="true" style="vertical-align: -0.505ex; width:20.145ex; height:2.843ex;" alt="{\displaystyle \therefore {AC}^{2}={AB}\times {AH}}"> <img src="
In triangle BHC and triangle ACB, ∠BHC=∠ACB as each is a right angle. ∠HBC=∠CBA as they are common angles at vertex B. Thus triangle BHC is similar to triangle BCA by AA test.
Thus, https://wikimedia.org/api/rest_v1/media/math/render/svg/5204aecd8df788ba689571a2ed9b9b07379dfc9f" class="mwe-math-fallback-image-inline mw-math-element" aria-hidden="true" style="vertical-align: -2.005ex; width:12.181ex; height:5.509ex;" alt="{\displaystyle {\frac {BC}{AB}}={\frac {BH}{BC}}}"> <img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/610b14b3148451604a2d4167a94187f4f1de962d" class="mwe-math-fallback-image-inline mw-math-element" aria-hidden="true" style="vertical-align: -0.505ex; width:20.187ex; height:2.843ex;" alt="{\displaystyle \therefore {BC}^{2}={AB}\times {BH}}"> <img src="
https://wikimedia.org/api/rest_v1/media/math/render/svg/38697cb9cef5c4989f32c753701dde1d289c25cb" class="mwe-math-fallback-image-inline mw-math-element" aria-hidden="true" style="vertical-align: -0.838ex; width:44.379ex; height:3.176ex;" alt="{\displaystyle \therefore {AC}^{2}+{BC}^{2}=({AB}\times {AH})+({AB}\times {BH})}"> <img src="
https://wikimedia.org/api/rest_v1/media/math/render/svg/b5c333b293240a928ac25b050a8acfe8cf2be79a" class="mwe-math-fallback-image-inline mw-math-element" aria-hidden="true" style="vertical-align: -0.838ex; width:36.17ex; height:3.176ex;" alt="{\displaystyle \therefore {AC}^{2}+{BC}^{2}={AB}\times ({AH}+{BH})}"> <img src="
https://wikimedia.org/api/rest_v1/media/math/render/svg/8b784bac3165373a60be56b542c2dc1ab1d68bba" class="mwe-math-fallback-image-inline mw-math-element" aria-hidden="true" style="vertical-align: -0.505ex; width:27.34ex; height:2.843ex;" alt="{\displaystyle \therefore {AC}^{2}+{BC}^{2}={AB}\times {AB}}"> <img src="
https://wikimedia.org/api/rest_v1/media/math/render/svg/9fe8d84a0c2bd91a5b939d779312869fef5b742b" class="mwe-math-fallback-image-inline mw-math-element" aria-hidden="true" style="vertical-align: -0.505ex; width:22.023ex; height:2.843ex;" alt="{\displaystyle \therefore {AC}^{2}+{BC}^{2}={AB}^{2}}"> which is the Pythagoras theorem. <img src="
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current | 14:12, 9 January 2017 | 512 × 385 (2 KB) | 127.0.0.1 (talk) | Simplified version of <a href="https://en.wikipedia.org/wiki/Pythagorean_theorem#Proof_using_similar_triangles" class="extiw" title="w:Pythagorean theorem">similar triangles proof</a> for <a href="//commons.wikimedia.org/wiki/Category:Pythagorean_theorem" title="Category:Pythagorean theorem">Pythagoras' theorem.</a><br><p>In triangle ACB, angle ACB is the right angle. CH is a perpendicular on hypotenuse AB of triangle ACB.<br> In triangle AHC and triangle ACB, ∠AHC=∠ACB as each is a right angle. ∠HAC=∠CAB as they are common angles at vertex A. Thus triangle AHC is similar to triangle ACB by <a href="https://en.wikipedia.org/wiki/Similarity_(geometry)#Angle.2Fside_similarities_for_triangles" class="extiw" title="w:Similarity (geometry)">AA test.</a><br> Thus, <span><span class="mwe-math-mathml-inline mwe-math-mathml-a11y mw-math-element" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><mrow class="MJX-TeXAtom-ORD"><mfrac><mrow><mi>A</mi><mi>C</mi></mrow><mrow><mi>A</mi><mi>B</mi></mrow></mfrac></mrow><mo>=</mo><mrow class="MJX-TeXAtom-ORD"><mfrac><mrow><mi>A</mi><mi>H</mi></mrow><mrow><mi>A</mi><mi>C</mi></mrow></mfrac></mrow></mstyle></mrow><annotation encoding="application/x-tex">{\displaystyle {\frac {AC}{AB}}={\frac {AH}{AC}}}</annotation></semantics></math></span><img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/edcf1ae6eaca6325cc257f26561db4c9aa6c499a" class="mwe-math-fallback-image-inline mw-math-element" aria-hidden="true" style="vertical-align: -2.005ex; width:12.139ex; height:5.509ex;" alt="{\displaystyle {\frac {AC}{AB}}={\frac {AH}{AC}}}"></span> <span><span class="mwe-math-mathml-inline mwe-math-mathml-a11y mw-math-element" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><mo>∴<!-- ∴ --></mo><msup><mrow class="MJX-TeXAtom-ORD"><mi>A</mi><mi>C</mi></mrow><mrow class="MJX-TeXAtom-ORD"><mn>2</mn></mrow></msup><mo>=</mo><mrow class="MJX-TeXAtom-ORD"><mi>A</mi><mi>B</mi></mrow><mo>×<!-- × --></mo><mrow class="MJX-TeXAtom-ORD"><mi>A</mi><mi>H</mi></mrow></mstyle></mrow><annotation encoding="application/x-tex">{\displaystyle \therefore {AC}^{2}={AB}\times {AH}}</annotation></semantics></math></span><img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/89edfb090cfb5ac5f9e4465566602cb0526af347" class="mwe-math-fallback-image-inline mw-math-element" aria-hidden="true" style="vertical-align: -0.505ex; width:20.145ex; height:2.843ex;" alt="{\displaystyle \therefore {AC}^{2}={AB}\times {AH}}"></span> </p> <p>In triangle BHC and triangle ACB, ∠BHC=∠ACB as each is a right angle. ∠HBC=∠CBA as they are common angles at vertex B. Thus triangle BHC is similar to triangle BCA by AA test.<br> Thus, <span><span class="mwe-math-mathml-inline mwe-math-mathml-a11y mw-math-element" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><mrow class="MJX-TeXAtom-ORD"><mfrac><mrow><mi>B</mi><mi>C</mi></mrow><mrow><mi>A</mi><mi>B</mi></mrow></mfrac></mrow><mo>=</mo><mrow class="MJX-TeXAtom-ORD"><mfrac><mrow><mi>B</mi><mi>H</mi></mrow><mrow><mi>B</mi><mi>C</mi></mrow></mfrac></mrow></mstyle></mrow><annotation encoding="application/x-tex">{\displaystyle {\frac {BC}{AB}}={\frac {BH}{BC}}}</annotation></semantics></math></span><img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/5204aecd8df788ba689571a2ed9b9b07379dfc9f" class="mwe-math-fallback-image-inline mw-math-element" aria-hidden="true" style="vertical-align: -2.005ex; width:12.181ex; height:5.509ex;" alt="{\displaystyle {\frac {BC}{AB}}={\frac {BH}{BC}}}"></span> <span><span class="mwe-math-mathml-inline mwe-math-mathml-a11y mw-math-element" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><mo>∴<!-- ∴ --></mo><msup><mrow class="MJX-TeXAtom-ORD"><mi>B</mi><mi>C</mi></mrow><mrow class="MJX-TeXAtom-ORD"><mn>2</mn></mrow></msup><mo>=</mo><mrow class="MJX-TeXAtom-ORD"><mi>A</mi><mi>B</mi></mrow><mo>×<!-- × --></mo><mrow class="MJX-TeXAtom-ORD"><mi>B</mi><mi>H</mi></mrow></mstyle></mrow><annotation encoding="application/x-tex">{\displaystyle \therefore {BC}^{2}={AB}\times {BH}}</annotation></semantics></math></span><img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/610b14b3148451604a2d4167a94187f4f1de962d" class="mwe-math-fallback-image-inline mw-math-element" aria-hidden="true" style="vertical-align: -0.505ex; width:20.187ex; height:2.843ex;" alt="{\displaystyle \therefore {BC}^{2}={AB}\times {BH}}"></span> <br><br></p> <p><span><span class="mwe-math-mathml-inline mwe-math-mathml-a11y mw-math-element" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><mo>∴<!-- ∴ --></mo><msup><mrow class="MJX-TeXAtom-ORD"><mi>A</mi><mi>C</mi></mrow><mrow class="MJX-TeXAtom-ORD"><mn>2</mn></mrow></msup><mo>+</mo><msup><mrow class="MJX-TeXAtom-ORD"><mi>B</mi><mi>C</mi></mrow><mrow class="MJX-TeXAtom-ORD"><mn>2</mn></mrow></msup><mo>=</mo><mo stretchy="false">(</mo><mrow class="MJX-TeXAtom-ORD"><mi>A</mi><mi>B</mi></mrow><mo>×<!-- × --></mo><mrow class="MJX-TeXAtom-ORD"><mi>A</mi><mi>H</mi></mrow><mo stretchy="false">)</mo><mo>+</mo><mo stretchy="false">(</mo><mrow class="MJX-TeXAtom-ORD"><mi>A</mi><mi>B</mi></mrow><mo>×<!-- × --></mo><mrow class="MJX-TeXAtom-ORD"><mi>B</mi><mi>H</mi></mrow><mo stretchy="false">)</mo></mstyle></mrow><annotation encoding="application/x-tex">{\displaystyle \therefore {AC}^{2}+{BC}^{2}=({AB}\times {AH})+({AB}\times {BH})}</annotation></semantics></math></span><img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/38697cb9cef5c4989f32c753701dde1d289c25cb" class="mwe-math-fallback-image-inline mw-math-element" aria-hidden="true" style="vertical-align: -0.838ex; width:44.379ex; height:3.176ex;" alt="{\displaystyle \therefore {AC}^{2}+{BC}^{2}=({AB}\times {AH})+({AB}\times {BH})}"></span><br></p> <p><span><span class="mwe-math-mathml-inline mwe-math-mathml-a11y mw-math-element" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><mo>∴<!-- ∴ --></mo><msup><mrow class="MJX-TeXAtom-ORD"><mi>A</mi><mi>C</mi></mrow><mrow class="MJX-TeXAtom-ORD"><mn>2</mn></mrow></msup><mo>+</mo><msup><mrow class="MJX-TeXAtom-ORD"><mi>B</mi><mi>C</mi></mrow><mrow class="MJX-TeXAtom-ORD"><mn>2</mn></mrow></msup><mo>=</mo><mrow class="MJX-TeXAtom-ORD"><mi>A</mi><mi>B</mi></mrow><mo>×<!-- × --></mo><mo stretchy="false">(</mo><mrow class="MJX-TeXAtom-ORD"><mi>A</mi><mi>H</mi></mrow><mo>+</mo><mrow class="MJX-TeXAtom-ORD"><mi>B</mi><mi>H</mi></mrow><mo stretchy="false">)</mo></mstyle></mrow><annotation encoding="application/x-tex">{\displaystyle \therefore {AC}^{2}+{BC}^{2}={AB}\times ({AH}+{BH})}</annotation></semantics></math></span><img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/b5c333b293240a928ac25b050a8acfe8cf2be79a" class="mwe-math-fallback-image-inline mw-math-element" aria-hidden="true" style="vertical-align: -0.838ex; width:36.17ex; height:3.176ex;" alt="{\displaystyle \therefore {AC}^{2}+{BC}^{2}={AB}\times ({AH}+{BH})}"></span><br></p> <p><span><span class="mwe-math-mathml-inline mwe-math-mathml-a11y mw-math-element" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><mo>∴<!-- ∴ --></mo><msup><mrow class="MJX-TeXAtom-ORD"><mi>A</mi><mi>C</mi></mrow><mrow class="MJX-TeXAtom-ORD"><mn>2</mn></mrow></msup><mo>+</mo><msup><mrow class="MJX-TeXAtom-ORD"><mi>B</mi><mi>C</mi></mrow><mrow class="MJX-TeXAtom-ORD"><mn>2</mn></mrow></msup><mo>=</mo><mrow class="MJX-TeXAtom-ORD"><mi>A</mi><mi>B</mi></mrow><mo>×<!-- × --></mo><mrow class="MJX-TeXAtom-ORD"><mi>A</mi><mi>B</mi></mrow></mstyle></mrow><annotation encoding="application/x-tex">{\displaystyle \therefore {AC}^{2}+{BC}^{2}={AB}\times {AB}}</annotation></semantics></math></span><img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/8b784bac3165373a60be56b542c2dc1ab1d68bba" class="mwe-math-fallback-image-inline mw-math-element" aria-hidden="true" style="vertical-align: -0.505ex; width:27.34ex; height:2.843ex;" alt="{\displaystyle \therefore {AC}^{2}+{BC}^{2}={AB}\times {AB}}"></span><br></p> <span><span class="mwe-math-mathml-inline mwe-math-mathml-a11y mw-math-element" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><mo>∴<!-- ∴ --></mo><msup><mrow class="MJX-TeXAtom-ORD"><mi>A</mi><mi>C</mi></mrow><mrow class="MJX-TeXAtom-ORD"><mn>2</mn></mrow></msup><mo>+</mo><msup><mrow class="MJX-TeXAtom-ORD"><mi>B</mi><mi>C</mi></mrow><mrow class="MJX-TeXAtom-ORD"><mn>2</mn></mrow></msup><mo>=</mo><msup><mrow class="MJX-TeXAtom-ORD"><mi>A</mi><mi>B</mi></mrow><mrow class="MJX-TeXAtom-ORD"><mn>2</mn></mrow></msup></mstyle></mrow><annotation encoding="application/x-tex">{\displaystyle \therefore {AC}^{2}+{BC}^{2}={AB}^{2}}</annotation></semantics></math></span><img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/9fe8d84a0c2bd91a5b939d779312869fef5b742b" class="mwe-math-fallback-image-inline mw-math-element" aria-hidden="true" style="vertical-align: -0.505ex; width:22.023ex; height:2.843ex;" alt="{\displaystyle \therefore {AC}^{2}+{BC}^{2}={AB}^{2}}"></span> which is the Pythagoras theorem. |
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