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Question
In an isosceles triangle ABC; AB = AC and D is the point on BC produced.
Prove that: AD2 = AC2 + BD.CD.
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Solution
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)
From equation (ii)
AD2 = AC2 - EC2 + (EC + CD)2
AD2 = AC2 - EC2 + EC2 + CD2 + 2EC·CD
AD2 = AC2 + CD (CD + 2EC)
AD2 = AC2 + CD (CD + BC)
AD2 = AC2 + CD·BD
Hence, proved.
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