Page 7 - ln1

0
∠ C = 180 - ∠ A - ∠ B
By substituting the values in the above equation
0
0
0
∠ C = 180 - 70 - 60
0
0
∠ C = 180 - 130
0
∠ C = 50
Consider △ BCD
We know that ∠ CBD is the exterior angle of ∠ ABC
So we get∠ CBD = ∠ DAC + ∠ ACB
By substituting the values in the above equation
0
0
∠ CBD = 70 + 50
0
By addition∠ CBD = 120
It is given that BC = BD
So we can write it as∠ BCD = ∠ BDC
Based on the sum property of the triangle
0
∠ BCD + ∠ BDC + ∠ CBD = 180
0
So we get∠ BCD + ∠ BDC = 180 - ∠ CBD
By substituting values in the above equation
0
0
∠ BCD + ∠ BDC = 180 - 120
0
∠ BCD + ∠ BDC = 60
0
It can be written as2 ∠ BCD = 60
0
By division∠ BCD = ∠ BDC = 30
In △ ACD
0
0
It is given that ∠ A = 70 and ∠ B = 60
We can write it as∠ ACD = ∠ ACB + ∠ BCD
By substituting the values we get
0
0
∠ ACD = 50 + 30
0
∠ ACD = 80
So we get to know that ∠ ACD is the greatest angle and the side opposite to it i.e. AD is
the longest side.
Therefore, it is proved that AD > CD
We know that ∠ BDC is the smallest angle and the side opposite to it i.e. AC is the
shortest side.
Therefore, it is proved that AD > AC.

Question 5.
If O is a point within △ABC, show that
(i) AB + AC > OB + OC
(ii) AB + BC + CA > OA + OB + OC
(iii) OA + OB + OC > ½ (AB + BC + CA)
Solution:

(i) It is given that O is a point within △ ABC
Consider △ ABC