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Question
In triangle ABC, angle A = 90o, CA = AB and D is the point on AB produced.
Prove that DC2 - BD2 = 2AB.AD.
Solution
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 ΔACD and applying Pythagoras theorem we get,
CD2 = AC2 + AD2
CD2 = AC2 + ( AB + BD )2 ....[ ∵ AD = AB + BD ]
CD2 = AC2 + AB2 + BD2 + 2AB.BD ...(i)
Similarly, in ΔABC,
BC2 = AC2 + AB2
BC2 = 2AB2 ...[ AB = AC ]
AB2 = `1/2`BC2 ...(ii)
Putting, AB2 from (ii) in (i), We get,
CD2 = AC2 + `1/2`BC2 + BD2 + 2AB . BD
CD2 - BD2 = AB2 + AB2 + 2AB . ( AD - AB )
CD2 - BD2 = AB2 + AB2 + 2AB . AD - 2AB2
CD2 - BD2 = 2AB . AD
DC2 - BD2 = 2AB . AD
Hence Proved.
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