A new proof .

$$Triangle ABC\quad be\quad a\quad right\quad triangle\quad with\quad \angle ABC\quad =90.\quad BD\quad be\quad the\quad perpendicular\quad on\quad AC\quad .\\ \quad \\ Then,\triangle ABD\sim \triangle ABC\quad and\quad \triangle BDC\quad \sim \triangle ABC.\\ \frac { ar(\triangle BDC) }{ ar(\triangle ABC)\quad } =\frac { { BC }^{ 2 } }{ { AC }^{ 2 } } \quad \quad \quad -{ eq }uation\quad 1\\ \frac { { ar(\triangle ADB) } }{ { ar(\triangle }ABC) } =\frac { { AB }^{ 2 } }{ { AC }^{ 2 } } \quad \quad \quad \quad -equation\quad 2\\ Adding\quad 1\& 2\quad we\quad get,\\ \frac { ar(\triangle ADB)+ar(\triangle BDC) }{ ar(\triangle ABC) } =\frac { { { BC }^{ 2 }+{ AB }^{ 2 } } }{ { AC }^{ 2 } } \\ \frac { ar(\triangle ABC) }{ ar(\triangle ABC) } =\frac { { { BC }^{ 2 }+{ AB }^{ 2 } } }{ { AC }^{ 2 } } \\ 1=\frac { { { BC }^{ 2 }+{ AB }^{ 2 } } }{ { AC }^{ 2 } } \\ { { AC }^{ 2 } }={ BC }^{ 2 }+{ AB }^{ 2 }\\ \\ Pythgoras\quad theorem\quad proved!$$

Note by Aryan Saxena
2 years, 11 months ago

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