show that in a right triangle the sqare of the hypotenuse is equal to the sum of the squares of other two sides
Answers
Answer:
Step-by-step explanation:
Given:
A right angled ∆ABC, right angled at B
To Prove- AC²=AB²+BC²
Construction: draw perpendicular BD onto the side AC .
Proof:
We know that if a perpendicular is drawn from the vertex of a right angle of a right angled triangle to the hypotenuse, than triangles on both sides of the perpendicular are similar to the whole triangle and to each other.
We have
△ADB∼△ABC. (by AA similarity)
Therefore, AD/ AB=AB/AC
(In similar Triangles corresponding sides are proportional)
AB²=AD×AC……..(1)
Also, △BDC∼△ABC
Therefore, CD/BC=BC/AC
(in similar Triangles corresponding sides are proportional)
Or, BC²=CD×AC……..(2)
Adding the equations (1) and (2) we get,
AB²+BC²=AD×AC+CD×AC
AB²+BC²=AC(AD+CD)
( From the figure AD + CD = AC)
AB²+BC²=AC . AC
Therefore, AC²=AB²+BC²
This theroem is known as Pythagoras theroem...
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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 .
