Q22 of 95 Page 5

In an isosceles ΔABC, AB = AC and BD ⊥ AC. Prove that BD2 — CD2 = 2CD AD.


Given: AB = AC and BD AC


To Prove: BD2 – CD2 = 2CD × AD


In ∆BDC, using Pythagoras theorem we have,


(Perpendicular)2 + (Base)2 = (Hypotenuse)2


⇒ (BD)2 + (CD)2 = (BC)2 …(i)


In ∆BDA, using Pythagoras theorem we have,


(Perpendicular)2 + (Base)2 = (Hypotenuse)2


⇒ (BD)2 + (AD)2 = (AB)2


⇒ (BD)2 + (AD)2 = (AC)2 [∵ AB = AC]


Multiply this eq. by 2, we get


⇒ 2(BD)2 + 2(AD)2 = 2(AC)2 …(ii)


Subtracting Eq. (ii) from (i), we get


⇒ CD2 – BD2 = BC2 – 2 AC2 + 2 AD2


= BC2 – 2 (AD +CD)2 + 2 AD2


= BC2 – 2 CD2 – 4 AD × CD


= BD2 + CD2 – 2 CD2 – 4 AD × CD


= BD2 – CD2 – 4 AD × CD


⇒ CD2 – BD2 –BD2 +CD2 = –4AD × CD


⇒ –2(BD2 – CD2) = –4AD × CD


⇒ BD2 – CD2 = 2CD × AD


Hence Proved


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