Question

prove pythagoras theorem

Answer 1

In this picture triangle ABC has an right angle at A,
so BC is the hypotenuse

To prove : BC* = CA* + AB*i.e. a*=b*+ c*,. (*=square)
in ABC and DEB,
CA = BD (by construction)
AB=DE(by construction)
L A =L D (each 90 degree) (L =angle)
so,ABC is congurent to DEB (SAS)
BC=BE (c.p.c.t.)
BE=a ( therefore, BC=a)
L ACB = L DBE (c.p.c.t.)

In ABC,L A +L ABC + LACB = 180
LA = 90
LABC + LDEB =90
since sum of angles at a point on one side of straight line is 180,
LABC + LDBE + LCBE =180
so, LCBE = 90
CBE is a right angle triangle at B

Now, LA +LD =90+90=180
AC is parallel to DE (sum of co-int. angles=180)
CADE is a trapezium.

From figure,
area of CADE = area CAB + area BDE + area CBE
1/2(CA+ED) x AD =1/2 CA x AB +1/2 BD x DE +1/2 CB x EB
(b+c)(c+b) = bc + bc + a x a
(b+c)*=2bc + a*
b* +c* = a*
CA* + AB* = BC*
proved Pythagoras theorum......
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Answer 2

Step-by-step explanation:

Pythagoras' theorem :-

→ In a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.

Step-by-step explanation:

It's prove :-

➡ Given :-

→ A △ABC in which ∠ABC = 90° .

➡To prove :-

→ AC² = AB² + BC² .

➡ Construction :-

→ Draw BD ⊥ AC .

➡ Proof :-

In △ADB and △ABC , we have

∠A = ∠A ( common ) .

∠ADB = ∠ABC [ each equal to 90° ] .

∴ △ADB ∼ △ABC [ By AA-similarity ] .

⇒ AD/AB = AB/AC .

⇒ AB² = AD × AC ............(1) .

In △BDC and △ABC , we have

∠C = ∠C ( common ) .

∠BDC = ∠ABC [ each equal to 90° ] .

∴ △BDC ∼ △ABC [ By AA-similarity ] .

⇒ DC/BC = BC/AC .

⇒ BC² = DC × AC. ............(2) .

Add in equation (1) and (2) , we get

⇒ AB² + BC² = AD × AC + DC × AC .

⇒ AB² + BC² = AC( AD + DC ) .

⇒ AB² + BC² = AC × AC .

$\huge \green{ \boxed{ \sf \therefore AC^2 = AB^2 + BC^2 }}$

Hence, it is proved.