prove that in a right triangle the square of the hypotenuse is equal to the sum of these square of the other 2 side
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Step-by-step explanation:
Given:- A right triangle ABC right angled at B.
To prove:- AC
2
=AB
2
+BC
2
Construction:- Draw BD⊥AC
Proof:-
In △ABC and △ABD
∠ABC=∠ADB(Each 90°)
∠A=∠A(Common)
∴△ABC∼△ABD(By AA)
AC
AB
=
AB
AD
(∵Sides of similar triangles are proportional)
⇒AB
2
=AD⋅AC.....(1)
Similarly, in △ABC and △BCD
∠ABC=∠BDC(Each 90°)
∠C=∠C(Common)
∴△ABC∼△BCD(By AA)
∴
BC
DC
=
AC
BC
⇒BC
2
=DC⋅AC.....(2)
Adding equation (1)&(2), we have
AB
2
+BC
2
=AD⋅AC+DC.AC
⇒AB
2
+BC
2
=AC(AD+DC)
⇒AB
2
+BC
2
=AC
2
Hence proved.
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