prove that In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
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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.....
Step-by-step explanation:
➠ AnSwer :-
➝ 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
→ AD × AC = AB²
In ∆BDC and ∆ABC, we have
→ ∠C=∠C. (common)
→ ∠BDC = ∠ABC [each equal to 90°]
→ ∆BDC ~ ∆ABC [by AA-similarity]
→ DC/BC = BC/AC
→ DC × AC = BC²
From (i) and (ii), we get
→ AD × AC+DC × AC = AB²+ BC²
→ (AD + DC) × AC = AB² + BC²
→ AC × AC = AB² + BC² = AC²= AB² + BC²
Hence,proved
