Question

In a right triangle, prove that the square on the hypotenuse is equal to sum of the squares on the other two sides

Answer 1

Figure is in the attachment


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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Hope this will help you.....

Answer 2

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 .


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


Hence, it is proved.