Mathematics
In an isosceles triangle ABC, AB = AC and D is a point on BC produced. Prove that AD2 = AC2 + BD × CD.
Pythagoras Theorem
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Answer
Draw AP ⊥ BC.
APD is right triangle
By pythagoras theorem we get,
⇒ AD2 = AP2 + PD2
⇒ AD2 = AP2 + (PC + CD)2
⇒ AD2 = AP2 + PC2 + CD2 + 2PC.CD ……(i)
APC is right triangle,
By pythagoras theorem we get,
⇒ AC2 = AP2 + PC2 ……(ii)
Substituting the value of AP2 + PC2 from (ii) in (i) we get,
⇒ AD2 = AC2 + CD2 + 2PC.CD …….(iii)
Since, ABC is an isosceles triangle,
⇒ PC = [∵ altitude to the base of an isosceles triangle bisects the base]
⇒ AD2 = AC2 + CD2 +
⇒ AD2 = AC2 + CD2 + BC.CD
⇒ AD2 = AC2 + CD(CD + BC)
From figure, CD + BC = BD
⇒ AD2 = AC2 + CD.BD
Hence, proved that AD2 = AC2 + BD × CD.
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