According to the exterior angle theorem, the exterior angle of a triangle is equal to the sum of the two opposite interior angles
The measure of ∠B is 44°
Reason:
The known parameters are;
The point D is on side [tex]\overline{AB}[/tex] of triangle ΔABC
ΔABC ~ ΔACD,
∠A = 48°
BC = CD
Required:
The measure of angle ∠B
Solution:
ΔBDC is an isosceles triangle
Let x, represent ∠B
Therefore, ∠BDC = x, by base angles of an isosceles triangle
Given that ΔABC ~ ΔACD, and ΔBDC is an isosceles triangle, we have;
∠ADC = 2·x by exterior angle of a triangle theorem
Given ΔABC ~ ΔACD, ∠A = ∠A, and ∠ADC = 2·x, therefore, ∠ACD = x
Which gives;
48° + x + 2·x = 180° sum of angles in a triangle theorem
[tex]x = \dfrac{180^{\circ} - 48^{\circ}}{3} = 44^{\circ}[/tex]
∠B = x = 44°
The measure of ∠B = 44°
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