Question

isosceles triangle ABC with AB=AC and BD perpendicular to AC ,prove that BC^2=2AC.CD

Answer 1

here's your ans
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]
Proved
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Answer 2

Answer:

BC^2 = 2AC.CD

Step-by-step explanation:

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]

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