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# Look at the Fig. 5.3. Show that length AH > sum of lengths of AB + BC + CD. Solve using Euclid’s axiom

**Solution:**

The figure represents the points A, B, C, D, E, F, G and H on the number line.

We have to show that the length AH > sum of lengths of AB + BC + CD

From the figure,

AH = AB + BC + CD + DE + EF + GH ----------- (1)

AB, BC, CD, DE, EF and GH are the parts of AH

Similarly, AB + BC + CD = AD --------------------- (2)

So, AB, BC and CD are the parts of AD

Using Euclid’s axiom,

The whole is greater than the part.

From (1) and (2), we observe that

AD is a part of AH

Length of AH = length of AD + DE + EF + GH

Therefore, length AH > sum of lengths of AB + BC + CD

**✦ Try This:** Two lines are respectively perpendicular to two parallel lines. Show that they are parallel to each other.

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

**NCERT Exemplar Class 9 Maths Exercise 5.3 Problem 3**

## Look at the Fig. 5.3. Show that length AH > sum of lengths of AB + BC + CD. Solve using Euclid’s axiom

**Summary:**

From the fig. 5.3 it is shown that the length AH is greater than the sum of lengths of AB + BC + CD by using Euclid’s axiom

**☛ Related Questions:**

- In Fig.5.4, we have AB = BC, BX = BY. Show that AX = CY. Solve using Euclid’s axiom
- In Fig.5.5, we have X and Y are the mid-points of AC and BC and AX = CY. Show that AC = BC. Solve us . . . .
- In Fig.5.6, we have BX = 1/2 AB, BY = 1/2 BC and AB = BC. Show that BX = BY. Solve using Euclid’s ax . . . .

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