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प्रश्न
In an isosceles triangle ABC; AB = AC and D is the point on BC produced.
Prove that: AD2 = AC2 + BD.CD.
उत्तर
In an isosceles triangle ABC; AB = AC and
D is the point on BC produced.
Construct AE perpendicular BC.
Pythagoras theorem states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.
We consider the rt. angled ΔAED and applying Pythagoras theorem we get,
AD2 = AE2 + ED2
AD2 = AE2 + ( EC + CD )2 ....(i)[ ∵ ED = EC + CD ]
Similarly, in ΔAEC,
AC2 = AE2 + EC2
AE2 = AC2 - EC2 ....(ii)
Putting AE2 = AC2 - EC2 in (i), We get,
AD2 = AC2 - EC2 + ( EC + CD )2
= AC2 + CD( CD + 2EC )
AD2 = AC2 + BD.CD .....[ ∵ 2EC + CD = BD ]
Hence proved.
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