D is any point on side AC of a ∆ ABC with AB = AC. Show that CD < BD.

Solution:

Given, ABC is a triangle

D is any point on side AC of the triangle

AB = AC

We have to show that CD < BD

We know that the angles opposite to the equal sides are equal.

∠ABC = ∠ACB ------------ (1)

Considering triangles ABC and DBC,

∠B = common angle

Since ∠DBC is a internal angle of ∠B

So, ∠ABC > ∠DBC

From (1), ∠ACB > ∠DBC

We know that in a triangle a side opposite to a greater angle is longer.

BD > CD

Therefore, CD < BD

✦ Try This: In triangle ABC, DE || BC. AD = x, DB = x - 2, AE = x + 2 and EC = x - 1, find the value x.

☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 7

NCERT Exemplar Class 9 Maths Exercise 7.3 Problem 7

D is any point on side AC of a ∆ ABC with AB = AC. Show that CD < BD

Summary:

D is any point on side AC of a ∆ ABC with AB = AC. It is shown that CD < BD

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