In the Fig. 5.10, we have AC = DC, CB = CE. Show that AB = DE. Solve using Euclid’s axiom
Solution:
The figure represents two triangles ADC and CBE with common vertex C.
Given, AC = DC --------------------- (1)
Also, CB = CE ----------------------- (2)
We have to show that AB = DE.
By using Euclid’s axiom,
If equals are added to the equals, the wholes are equal.
On adding (1) and (2),
AC + CB = DC + CE
From the figure,
AC + CB = AB --------------------- (3)
Similarly, DC + EC = DE -------- (4)
From (3) and (4),
AB = DE
Therefore, it is proved that AB = DE
✦ Try This: In the figure, DE || QR and AP and BP are bisectors of ∠EAB and ∠RBA, respectively. Find ∠APB.
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 5
NCERT Exemplar Class 9 Maths Exercise 5.3 Problem 10
In the Fig. 5.10, we have AC = DC, CB = CE. Show that AB = DE. Solve using Euclid’s axiom
Summary:
The figure consists of two triangles with common vertex C. In the figure we have AC = DC, CB = CE. By using Euclid’s axiom, it is shown that AB = DE
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