In Fig. 6.2, ∠BAC = 90° and AD⟂BC. Then,
a. BD . CD = BC²
b. AB . AC = BC²
c. BD . CD = AD²
d. AB . AC = AD²

Solution:

Given, ∠BAC = 90°

Also, AD⟂BC

We know that a perpendicular drawn from the vertex of the right angle of a right triangle to its hypotenuse divides the triangle into two triangles which are similar to the whole triangle and to each other.

From the figure,

△BAD ~ △BCA

△BAD ~ △ACD

△CAD ~ △CBA

By using the property of similarity,

The corresponding sides of similar triangles are in proportion.

Since, △BAD ~ △ACD, the sides are proportional to each other.

BD/AD = AD/CD

On cross multiplication,

BD(CD) = AD(AD)

AD² = BD × CD

✦ Try This: In a Δ PQR, N is a point on PR, such that QN ⊥ PR. If PN . NR = QN2, then prove that ∠PQR = 90°

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 6

NCERT Exemplar Class 10 Maths Exercise 6.1 Problem 1

In Fig. 6.2, ∠BAC = 90° and AD⟂BC. Then, a. BD . CD = BC², b. AB . AC = BC², c. BD . CD = AD², d. AB . AC = AD²

Summary:

In Fig. 6.2, ∠BAC = 90° and AD⟂BC. Then, AD² = BD × CD

☛ Related Questions:

• The lengths of the diagonals of a rhombus are 16 cm and 12 cm. Then, the length of the side of the r . . . .
• If △ABC ~ △EDF and △ABC is not similar to △DEF, then which of the following is not true, a. BC . E . . . .
• If in two triangles ABC and PQR, AB/QR = BC/PR = CA/PQ, then, a. △PQR ~ △CAB, b. △PQR ~ △ABC, c. △CB . . . .