Mathematics
In triangle ABC, ∠B = 90° and D is the mid-point of BC. Prove that : AC2 = AD2 + 3CD2.
Pythagoras Theorem
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Answer
Given,
D is the mid-point of BC.
∴ CD = BD ……..(1)
By formula,
By pythagoras theorem,
⇒ (Hypotenuse)2 = (Perpendicular)2 + (Base)2
In right-angled △ ABC,
⇒ AC2 = AB2 + BC2
⇒ AC2 = AB2 + (BD + CD)2
⇒ AC2 = AB2 + (CD + CD)2 ……….[From equation (1)]
⇒ AC2 = AB2 + (2CD)2
⇒ AC2 = AB2 + 4 CD2 ………..(2)
In right-angled △ ABD,
⇒ AD2 = AB2 + BD2
⇒ AD2 = AB2 + CD2 [From equation (1)] …….(3)
Subtracting equation (3) from (2), we get :
⇒ AC2 - AD2 = AB2 + 4 CD2 - (AB2 + CD2)
⇒ AC2 - AD2 = AB2 - AB2 + 4 CD2 - CD2
⇒ AC2 - AD2 = 3 CD2
⇒ AC2 = AD2 + 3 CD2.
Hence, proved that AC2 = AD2 + 3 CD2.
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