Pythagoras Theorem with converse
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The converse of the Pythagorean Theorem is: If the square of the length of the longest side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle.
Step-by-step explanation:
Pythagoras' theorem :-
→ In a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.
Step-by-step explanation:
It's prove :-
➡ 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 .
Hence, it is proved.
Statement:
In a Triangle the square of longer side is equal to the sum of squares of the other two sides, then the triangle is a right angled triangle.
Given -
A Triangle ABC such that
BC² = AB² + AC²
To Prove -
Angle A = 90°
Construction -
Draw a ∆DEF such that AB = DE and AC = DF and Angle D = 90°
Proof -
In ∆ABC,
BC² = AB² + AC² - Given
In ∆ DEF
EF² = DE² + DF²
Therefore,
EF² = AB² + AC²
(Since AB = DE, AC = DF)
Therefore,
BC² = EF² ie - BC = EF
Now, In ∆ABC and ∆DEF
AB = DE - By Construction
AC = DF - By Construction
BC = EF
Therefore
∆ABC ≅ ∆DEF by SSS test.
Thus,
Angle A = Angle D - CPCT
But, Angle D = 90° ( As per construction)
Therefore
Angle A = 90°
Hence Proved!