Answer:
Given: In [tex]\triangle ABC[/tex], point D is on the side AB so that CA=CD.
Isosceles Triangle states:
* A triangle with two equal sides.
*The angles which are opposite to the equal sides are also equal.
Since, CA=CD
By definition of isosceles triangle,
[tex]\triangle CAD[/tex] is an isosceles triangle,
All the three angles within an isosceles triangle are acute. Acute angles are always less than 90 degrees.
therefore, [tex]\angle CAD=\angle CDA[/tex][tex]<90^{\circ}[/tex] ......[1]
Now, Consider [tex]\triangle CBD[/tex],
Exterior Angle Property state that an exterior angle of a triangle is equal to the sum of the opposite interior angles.
Therefore, by exterior angle property in [tex]\triangle CBD[/tex] we have ;
[tex]\angle CBD + \angle BCD = \angle CDA[/tex]
So, [tex]\angle CBD< \angle CDA[/tex]
From [1] we have:
[tex]\angle CBD < \angle CDA=\angle CAD <90^{\circ}[/tex]
Consider [tex]\triangle CAB[/tex],
Sine law states that an equation relating the lengths of the sides of a triangle to the sines of its angles.
Now, using Sine law in [tex]\triangle CAB[/tex],
[tex]\frac{\sin (\angle CAD)}{BC} =\frac{\sin (\angle CBD)}{AC}[/tex] ......[2]
Since,
[tex]\angle CBD < \angle CDA< 90^{\circ}[/tex]
so, [tex]\sin (\angle CBD)<\sin (\angle CAD)[/tex] ......[3]
From [2] and [3] we have;
[tex]AC<BC[/tex] or we can write it as [tex]BC>AC[/tex] hence Proved!
