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$D$ is on side $\overline {AB}$ of $\triangle ABC.$ We know $\triangle ABC \sim \triangle ACD$ and $\angle A =48^\circ.$ If $BC=BD,$ what is $\angle B$ in degrees?

Sagot :

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°

Learn more about angles in a triangle here:

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