Question

angle A of an isoceles triangle ABC is acute in which AB=AC, BD perpendicular AC. prove that BC square=2AC*CD

Answer 1

In Δ BCD by pythagoras theorem, we have

⇒ BC2 = BD2 + CD2 ... (1)

Again, in Δ ABD by pythagoras theorem

AB2 = BD2 + AD2

⇒ BD2 = AB2 - AD2

On putting value of BD2 in (1), we get

BC2 = AB2 - AD2 + CD2 

⇒ BC2 = AB2 - (AC - CD)2 + CD2

⇒ BC2 = AB2 - AC2 - CD2 + 2AC. CD + CD2

⇒ BC2 = 2AC. CD [Given, AB = AC. ⇒ AB2 = AC2]

[Hence proved]

Answer 2

In Δ BCD by pythagoras theorem, we have

⇒ BC2 = BD2 + CD2 ... (1)

Again, in Δ ABD by pythagoras theorem

AB2 = BD2 + AD2

⇒ BD2 = AB2 - AD2

On putting value of BD2 in (1), we get

BC2 = AB2 - AD2 + CD2 

⇒ BC2 = AB2 - (AC - CD)2 + CD2

⇒ BC2 = AB2 - AC2 - CD2 + 2AC. CD + CD2

⇒ BC2 = 2AC. CD [Given, AB = AC. ⇒ AB2 = AC2]