right angle triangle the square of the hypotenuse is equal to the sum of the squares of the other two sides prove
Answers
Answer:
given=IN A RIGHT ANGLED TRIANGLE ABC,RIGHT ANGLED AT B.
TO PROVE= AC²=AB²+BC²
CONSTRUCTION=DRAW BD PERPENDICULAR ON AC
PROOF=SINCE,ABC IS A RIGHT ANGLED TRIANGLE & BD PERPENDICULAR ON AC
TRIANGLE ADB~TRIANGLE ABC......~(1)
TRIANGLE BDC~TRIANGLE ABC.......~(2)
THEN,
FROM EQ.(1),WE HAVE
AD/DB=AB/AC
[SINCE,CORRESPONDING SIDES OF SIMILAR TRIANGLES ARE PROPOTIONAL]
AD.AC=AB².......~(3)
FROM EQ (2) WE HAVE
DC/BC=BC/AC
[SINCE,CORRESPONDUNG SIDES OF SIMILAR TRIANGLES ARE PROPOTIONAL]
DC.AC=BC²........~(4)
ON+EQS.(3)&(4),we get
AD.AC+DC.AC=AB²+BC²
(AD+DC).AC=AB²+BC²
AC.AC=AB²+BC²
AC²=AB²+BC²
HENCE PROVED
hope it will help you
Answer:
Given :
A right triangle ABC right angled at B.
To prove :
AC² = AB² + BC²
Construction :
Draw BD ⊥ AC
Proof :
In Δ ADB and Δ ABC
∠ A = ∠ A [ Common angle ]
∠ ADB = ∠ ABC [ Both are 90° ]
∴ Δ ADB Similar to Δ ABC [ By AA similarity ]
So , AD / AB = AB / AC [ Sides are proportional ]
= > AB² = AD . AC ... ( i )
Now in Δ BDC and Δ ABC
∠ C = ∠ C [ Common angle ]
∠ BDC = ∠ ABC [ Both are 90° ]
∴ Δ BDC Similar to Δ ABC [ By AA similarity ]
So , CD / BC = BC / AC
= > BC² = CD . AC ... ( ii )
Now adding both equation :
AB² + BC² = CD . AC + AD . AC
AB² + BC² = AC ( CD + AD )
AB² + BC² = AC² .
AC² = AB² + BC² .
Hence proved .
