# In Fig. 6.53, ABD is a triangle right angled at A and AC ⊥ BD. Show that

(i) AB^{2} = BC.BD

(ii) AC^{2} = BC.DC

(iii) AD^{2} = BD.CD

**Solution:**

We use the properties of right-angled triangles and perpendicular lines to solve the question given.

As we know if a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then triangles on both sides of the perpendicular are similar to the whole triangle and to each other.

(i) As we know if a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then triangles on both sides of the perpendicular line are similar to the whole triangle and to each other.

⇒ ΔBAD ∼ ΔBCA

⇒ AB/BC = BD/AB (Corresponding sides of similar triangle)

⇒ AB^{2} = BC.BD

(ii) As we know if a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then triangles on both sides of the perpendicular line are similar to the whole triangle and to each other.

⇒ ΔBCA ∼ ΔACD

⇒ AC/CD= BC/AC

⇒ AC^{2} = BC.DC (Corresponding sides of a similar triangle)

(iii) As we know that if a perpendicular line is drawn from the vertex of the right angle to the hypotenuse then triangles on both sides of the perpendicular line are similar to the whole triangle and to each other.

⇒ ΔBAD ∼ ΔACD

⇒ AD/CD = BD/AD

AD^{2} = BD.CD (Corresponding sides of a similar triangle)

**Video Solution:**

## In Fig. 6.53, ABD is a triangle right angled at A and AC ⊥ BD. Show that (i) AB^{2} = BC.BD (ii) AC^{2} = BC.DC (iii) AD^{2} = BD.CD

### NCERT Class 10 Maths Solutions - Chapter 6 Exercise 6.5 Question 3:

In Fig. 6.53, ABD is a triangle right angled at A and AC ⊥ BD. Show that (i) AB^{2} = BC.BD (ii) AC^{2} = BC.DC (iii) AD^{2} = BD.CD

It is proved tha(i) AB^{2} = BC.BD (ii) AC^{2} = BC.DC and (iii) AD^{2} = BD.CD