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prove pythagoras theorem.
Posted- 1720 days ago
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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:
AD ⊥ CB 
⇒∆ADB  is  a right triangle , right angled at D
From Pythagoras theorem, we have 
AC2= AD2+DB2
 
Also, AD ⊥ CB 
⇒∆ADC  is  a right triangle , right angled at D
From Pythagoras theorem, we have 
AC2  =  AD2 + DC2
 
⇒AC2 = AD2+(DB+BC) 2
 
⇒AC2= AD2+DB2 +BC2 +2 BC.BD 
⇒AC2= (AD2+ DB2)+BC2+2 BC.BD


⇒AC2= AB2+BC2+2 BC.BD

Thus, the theorum is proved.

 
Statement of Phythagoras Theorem: In right triangle the square of hypotanuse is equal to sum of squares of other two sides. Given = A right angled triangle, Right angled at B. To Prove that = AC2=AB2+BC2. Construction = Draw BD perpendicular to AC. Proof = In triangle ABD and triangle ABC. angle A = angle A (common) angle ABD = angle ABC (each 90*) By AA similarity criterian, triangle ABD~triangle ABC. AD/AB = AB/AC. AB2=AD.AC ------(1) Now, In triangle BDC and triangle ABC. angle C = angle C. (common) angle BDC = angle ABC (each 90*) By AA similarity criterian, triangle BDC~ triangle ABC. DC/BC = BC/AC BC2=AC.DC --------- (2) On adding (1) & (2). AB2+BC2 = AD.AC + AC.DC = AC (AD+DC). = AC X AC. = AC2. AB2+BC2 = AC2. Hence Proved.....!