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prove pythagoras theorem.
Posted- 1720 days ago

Let ABC represent a right triangle, with the right angle located at C, as shown on the figure. We draw the altitude from point C, and call H its intersection with the side AB. The new triangle ACH is similar to our 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. By a similar reasoning, the triangle CBH is also similar to ABC. The similarities lead to the two ratios:

$\frac{a}{c}=\frac{HB}{a} \mbox{ and } \frac{b}{c}=\frac{AH}{b}.\,$

These can be written as

$a^2=c\times HB \mbox{ and }b^2=c\times AH. \,$

Summing these two equalities, we obtain

$a^2+b^2=c\times HB+c\times AH=c\times(HB+AH)=c^2 .\,\!$

In other words, the Pythagorean theorem:

$a^2+b^2=c^2.\,\!$  PROVED...

In a right angled

AC square = AB square + BC square

AC = AB +BC

hence proved

In the given Figure, ABC is a triangle in which   ABC  >  90° and AD ⊥ CB
produced. Prove that
AC2 =  AB2 + BC2 + 2 BC . BD

Given : ABC is a triangle in which   ABC  >  90° and AD ⊥ CB produced.
To Prove: AC2 =  AB2 + BC2 + 2 BC . BD

Proof:
⇒∆ADB  is  a right triangle , right angled at D
From Pythagoras theorem, we have

⇒∆ADC  is  a right triangle , right angled at D
From Pythagoras theorem, we have