# 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 know that 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 each other and to the whole triangle.

(i) Here, ABD is a triangle right angled at A and AC ⊥ BD

⇒ ΔBAD ∼ ΔBCA

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

⇒ AB^{2} = BC.BD

(ii) ABD is a triangle right angled at A and AC ⊥ BD

⇒ ΔBCA ∼ ΔACD

⇒ AC/CD= BC/AC

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

(iii) ABD is a triangle right angled at A and AC ⊥ BD

⇒ ΔBAD ∼ ΔACD

⇒ AD/CD = BD/AD

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

**☛ Check: **NCERT Solutions Class 10 Maths Chapter 6

**Video Solution:**

## In Fig. 6.53, ABD is a triangle right angled at A and AC ⊥ BD. Show that (i) AB² = BC.BD (ii) AC² = BC.DC (iii) AD² = BD.CD

NCERT Class 10 Maths Solutions Chapter 6 Exercise 6.5 Question 3

**Summary:**

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

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- In Figure 6.54, O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that(i) OA2 + OB2 + OC2 - OD2 - OE2 - OF2 = AF2 + BD2 + CE2(ii) AF2 + BD2 + CE2 = AE2 + CD2 + BF2