The perpendicular bisector of side ab of ∆abc intersects the extension of side ac atd. find the measure of ∠abc if m∠cbd = 16° and m∠acb = 118°.
Accepted Solution
A:
Answer:[tex]\angle ABC=23^{\circ}[/tex]Step-by-step explanation:Given information: The perpendicular bisector of side AB of ∆ABC intersects the extension of side AC at D, m∠CBD = 16° and m∠ACB = 118°.Let the measure of ∠ABC is x°.[tex]\angle ABD=\angle ABC+\angle CBD[/tex][tex]\angle ABD=x+16[/tex]In triangle ABD, DM is perpendicular bisector of AB.In triangle ADM and BDM,[tex]AM\cong BM[/tex] (Definition of perpendicular bisector)[tex]\angle AMD\cong \angle BMD[/tex] (Definition of perpendicular bisector)[tex]DM\cong DM[/tex] (Reflection property)By SAS postulate,[tex]\triangle ADM\cong \triangle BDM[/tex][tex]\angle MAD\cong \triangle MBD[/tex] (CPCTC)[tex]\angle MAD=x+16[/tex][tex]\angle BAC=x+16[/tex]According to angle sum property of a triangle, the sum of interior angles of triangle is 180°. In triangle ABC[tex]\angle ABC+\angle ACB+\angle BAC=180[/tex][tex]x+118+(x+16)=180[/tex][tex]2x+134=180[/tex]Subtract 134 from both sides.[tex]2x=180-134[/tex][tex]2x=46[/tex]Divide both sides by 2.[tex]x=\frac{46}{2}[/tex][tex]x=23}[/tex]Therefore, the measure of ∠ABC is 23°.