In the Fig. 5.9, we have ∠ABC = ∠ACB, ∠3 = ∠4. Show that ∠1 = ∠2. Solve using Euclid’s axiom

Solution:

The figure represents two triangles ABC and BDC with common base BC.

Given, ∠ABC = ∠ACB ----------------- (1)

Also, ∠3 = ∠4 ---------------------------- (2)

We have to show that ∠1 = ∠2

From the figure,

∠ABC = ∠1 + ∠4

∠1 = ∠ABC - ∠4 ------------------------ (3)

∠ACB = ∠3 + ∠2

∠2 = ∠ACB - ∠3 ------------------------ (4)

By using Euclid’s axiom,

If equals be subtracted from equals, the remainders are equal.

From (3) and (4),

∠ABC - ∠4 = ∠ACB - ∠3

Therefore, ∠1 = ∠2

✦ Try This: AP and BQ are the bisectors of the two alternate interior angles formed by the intersection of a transversal t with parallel lines l and m (in the given figure). Show that AP || BQ.

☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 5

NCERT Exemplar Class 9 Maths Exercise 5.3 Problem 9

In the Fig. 5.9, we have ∠ABC = ∠ACB, ∠3 = ∠4. Show that ∠1 = ∠2. Solve using Euclid’s axiom

Summary:

The figure represents two triangles ABC and ADC with common base BC. We have ∠ABC = ∠ACB, ∠3 = ∠4. By using Euclid’s axiom, it is shown that ∠1 = ∠2

☛ Related Questions:

• In the Fig. 5.10, we have AC = DC, CB = CE. Show that AB = DE. Solve using Euclid’s axiom
• In the Fig. 5.11, if OX = 1/2 XY, PX = 1/2 XZ and OX = PX, show that XY = XZ..Solve using Euclid’s a . . . .
• In Fig.5.12, we have AB = BC, M is the mid-point of AB and N is the mid- point of BC. Show that AM = . . . .