Question

in △abc, the perpendicular bisector of side $overline{ab}$ intersects the extension of side $overline{ac}$ at point d. find the measure of ∠abc if m∠cbd = 16° and m∠acb = 118°. answer: m∠abc =

Explanation:

Step1: Find $\angle BCD$

$\angle BCD = 180^{\circ}-\angle ACB=180^{\circ} - 118^{\circ}=62^{\circ}$

Step2: Find $\angle BDC$ in $\triangle BCD$

In $\triangle BCD$, using the angle - sum property of a triangle ($\angle BCD+\angle CBD+\angle BDC = 180^{\circ}$), we have $\angle BDC=180^{\circ}-\angle BCD - \angle CBD$. Substituting the values, $\angle BDC = 180^{\circ}-62^{\circ}-16^{\circ}=102^{\circ}$.

Step3: Let the perpendicular bisector of $AB$ be $l$. Since $l$ is the perpendicular bisector of $AB$, $AD = BD$. So, $\angle A=\angle ABD$.

Let $\angle ABC=x$. Then $\angle ABD=x + 16^{\circ}$, so $\angle A=x + 16^{\circ}$.

Step4: Use the angle - sum property in $\triangle ABC$

In $\triangle ABC$, $\angle A+\angle ABC+\angle ACB=180^{\circ}$. Substituting $\angle A=x + 16^{\circ}$, $\angle ABC=x$ and $\angle ACB = 118^{\circ}$, we get $(x + 16^{\circ})+x+118^{\circ}=180^{\circ}$.
Combining like terms: $2x+134^{\circ}=180^{\circ}$.
Subtract $134^{\circ}$ from both sides: $2x=180^{\circ}-134^{\circ}=46^{\circ}$.
Divide both sides by 2: $x = 23^{\circ}$.

Answer:

$23^{\circ}$