Question

In a triangle ABC,N is a point on AC such that BN perpendicular AC. If BN2 = AN multiplyNC then prove that angle B =90 degree

Answer 1

It is given that, in ΔABC, N is point on AC such that, BN ⊥AC.

Also, BN²= AN × NC-------(1)

To prove : ∠B=90°

Solution:

In Right  Δ ANB

By, Pythagoras theorem

AB²= BN²+ AN²

AB²=AN × NC + AN²-----using (1)

AB²= AN ×[ NC +AN]

AB²=AN × AC----------(2)

In Right  Δ C NB

By, Pythagoras theorem

CB²= BN²+ CN²

CB²=AN × NC + C N²-----using (1)

CB²= C N ×[ NC +AN]

CB²=C N × AC----------(3)

Adding 2 and 3

AB²+CB²=C N × AC+AN × AC

               = AC×[CN + AN]

               = AC ×AC

  AB²+CB²=AC²,

So, by converse of pythagoras theorem , which states that if in a triangle sum of squares of two side is equal to square of third side then angle between two smaller sides is of 90°.

So, Angle between AB and BC is of 90°.

Means, ∠B=90°

Answer 2

Answer:

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