Question

State and prove converse of pythagoras.​

Answer 1

Answer:

Pythagoras Theorem

Statement: In a right angled triangle, the square of the hypotenuse is equal to the sum of squares of the other two sides.

Given: A right triangle ABC right angled at B.

To prove: AC2 = AB2 + BC2

Construction: Draw BD perpendicular to AC

Proof :

We know that: If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse, then triangles on both sides of the perpendicular are similar to the whole triangle and to each other.

ADB

ABC

So,

(Sides are proportional)

Or, AD.AC = AB2 ... (1)

Also,

BDC

ABC

So,

Or, CD. AC = BC2 ... (2)

Adding (1) and (2),

AD. AC + CD. AC = AB2 + BC2

AC (AD + CD) = AB2 + BC2

AC.AC = AB2 + BC2

AC2 = AB2 + BC2

Hence Proved.

Answer 2

Converse of Pythagoras Theorem:

In a triangle if the square of a side is equal to the sum of the squares of other two side then the angle opposite to the first side is a right angle.

Given:

$\sf{In\:ΔABC\:AB^2+BC^2=AC^2--(i)}$

To prove:

$\sf{\angle\: B = 90°}$

Construction:

$\sf{Construct\: another\: triangle\: ΔPQR\: such \:that}$

$\sf{PQ = AB, QR = BC \:and \:\angle\:Q= 90°}$

Proof:

$\sf{In \:ΔPQR}$,

$\sf{by\:Pythagoras\:theorem}$

$\sf{PQ^2+QR^2=PR^2}$

$\sf{AB^2+BC^2=PR^2--(ii)}$

$\sf{From\:(i)\:and\:(ii)}$

$\sf{AC^2=PR^2}$

$\sf{AC=PR}$

$\sf{In\:ΔABC\:and\:ΔPQR}$

$\sf{AB=BC(By\:construction)}$

$\sf{BC=QR(By\:construction}$

$\sf{AC=PR(Proved\:above)}$

$\sf{ΔABC ≅ΔPQR(By\:SSS\:criterion)}$

$\sf{By\:CPCT}$

$\sf\boxed{\angle\: B = 90°}$

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