Mathematics
In triangle ABC, angle A = 90°, CA = AB and D is a point on AB produced. Prove that :
DC2 - BD2 = 2AB.AD.
Pythagoras Theorem
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Answer
By formula,
By pythagoras theorem,
⇒ (Hypotenuse)2 = (Perpendicular)2 + Base2
In right angle triangle ACD,
By pythagoras theorem,
⇒ CD2 = AC2 + AD2
⇒ CD2 = AC2 + (AB + BD)2
⇒ CD2 = AC2 + AB2 + BD2 + 2.AB.BD ………(1)
In right angle triangle ABC,
By pythagoras theorem,
⇒ BC2 = AC2 + AB2
⇒ BC2 = AB2 + AB2 (Since, CA = AB)
⇒ BC2 = 2AB2
⇒ AB2 = BC2 ………(2)
Substituting value of AB2 from equation (2) in (1), we get :
⇒ CD2 = AC2 + + BD2 + 2.AB.BD
⇒ CD2 - BD2 = AB2 + + 2.AB.BD [As, AC = AB]
⇒ CD2 - BD2 = AB2 + AB2 + 2.AB.(AD - AB) [From equation (2)]
⇒ CD2 - BD2 = 2AB2 + 2.AB.AD - 2.AB2
⇒ CD2 - BD2 = 2.AB.AD
Hence, proved that CD2 - BD2 = 2.AB.AD
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Related Questions
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AC2 = 2.BC.DC.
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AD2 = AC2 + BD.CD
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BD2 - CD2 = 2CD × AD
In the following figure, AD is perpendicular to BC and D divides BC in the ratio 1 : 3.
Prove that : 2AC2 = 2AB2 + BC2.
