In the ∆ABC given below BC = 5cm
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In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
In △ACD
AD
2
=AC
2
+CD
2
.(1)
In △ACB
AB
2
−AC
2
=BC
2
..(2)
As CD=DB=
2
BC
. (3)
Substituting (3) in (1)
AD
2
=AC
2
+
4
BC
2
4AD
2
−4AC
2
=BC
2
. (4)
Subtracting (2) from (4)
4AD
2
−4AC
2
−(AB
2
−AC
2
)=BC
2
−BC
2
4AD
2
−4AC
2
−AB
2
+AC
2
=0
4AD
2
−3AC
2
=AB
2
Hence proved.
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