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  • Prove that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of other two sides.       <div

Prove that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of other two sides.

 

 
 
 
 
 

Answers (1)

In \triangleADB and \triangleABC

\angle BDA = \angle CBA\: \: \left \{ \because 90^{\circ} angles \right \}

\angle BAD = \angle BAC\: \: \left \{ \because common \right \}

\therefore \triangle ADB \sim \triangle ABC\: \: \left \{ \because AA \right \}

\therefore \frac{AD}{AB}= \frac{AB}{AC}   (from the theorem)

AB^2=AD.AC ........(1)

In \triangleBDC and \triangleABC,

\angle BDA = \angle ABC\: \: \left \{ \because 90^{\circ} angles \right \}

\angle BCD = \angle BCA\: \: \left \{ \because common \right \}

\therefore \triangle BDC \sim \triangle ABC\: \: \left \{ \because AA \right \}

\therefore \frac{CD}{BC}= \frac{BC}{AD}   (from the theorem)

BC^2=AC.CD ........(2)

 

Adding eqn(1) and eqn(2)

AB^2+BC^2=AD.AC+AC.CD

AB^2+BC^2=AC(AD+CD)

AB^2+BC^2=AC(AC)\: \: \: \left \{ \because AD+CD=AC \right \}

AB^2+BC^2=AC^2

Hence Proved.

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Safeer PP

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