In a right triangle, the square of the hypotenuse
equal to the sum ofthe squares of the other two sides
Answers
Answer:
Step-by-step explanation:
➡ 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 .
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 .