Question

Segment AN is the altitude to side BC in ΔABC. If AB = 3NC and AN = 2NC, prove that AC = BN. (Hint: Use variables in such problems. Let NC = x units and find the other lengths in terms of x.)

Answer 1

BN=AC=√5 x.

The proof is explained in step-by-step explaination.

Explanation:

Let NC=x. It is given that AB=3NC & AN=2NC

⇒ AB=3x & AN=2x

By applying Pythagoras theorem

In triangle ANC,

`AC^(2)=AN^(2)+NC^(2)`

⇒
`AC^(2) = (2x)^(2)+x^(2)`

⇒
`AC^(2)=4x^(2)+x^(2) =5x^(2)`

⇒
`AC=√(5)x` → (1)

Similarly, In triangle ABN,

`AB^(2)=AN^(2)+BN^(2)`

⇒
`(3x)^(2)=BN^(2)+x^(2)`

⇒
`9x^(2) = (BN)^(2)+4x^(2)`

⇒
`BN^(2)=5x^(2)`

⇒
`BN=√(5)x` → (2)

From eq (1) & (2), AC=BN

Answer 2

The proof is as follows :

Explanation:

Let NC = x

⇒ AB = 3x and AN = 2x

In Δ ABN, By using Pythagoras theorem,

AB² = BN² + AN²

⇒ BN² = AB² - AN²

⇒ BN² = (3x)² - (2x)²

⇒ BN² = 5x²

⇒ BN = x√5 .......................(1)

Now in ΔANC , Using Pythagoras theorem We have,

AC² = NC² + AN²

⇒ AC² = x² + (2x)²

⇒ AC² = 5x²

⇒ AC = x√5 ....................(2)

From equations (1) and (2) We get,

AC = BN , which is our required result