Question

in the isosceles triangle ABC, AB=AC and BD is the altitude through B. Prove that: BD^2-CD^2=2CD.AD​

Answer 1

Answer:

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