The perpendicular bisector of side AB of ∆ABC intersects the extension of side AC at D. Find the measure of ∠ABC if ∠CBD = 16° and m∠ACB = 118°
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Answer:
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°.
\angle ABD=\angle ABC+\angle CBD∠ABD=∠ABC+∠CBD
\angle ABD=x+16∠ABD=x+16
In triangle ABD, DM is perpendicular bisector of AB.
In triangle ADM and BDM,
AM\cong BMAM≅BM (Definition of perpendicular bisector)
\angle AMD\cong \angle BMD∠AMD≅∠BMD (Definition of perpendicular bisector)
DM\cong DMDM≅DM (Reflection property)
By SAS postulate,
\triangle ADM\cong \triangle BDM△ADM≅△BDM
\angle MAD\cong \triangle MBD∠MAD≅△MBD (CPCTC)
\angle MAD=x+16∠MAD=x+16
\angle BAC=x+16∠BAC=x+16
According to angle sum property of a triangle, the sum of interior angles of triangle is 180°.
In triangle ABC
\angle ABC+\angle ACB+\angle BAC=180∠ABC+∠ACB+∠BAC=180
x+118+(x+16)=180x+118+(x+16)=180
2x+134=1802x+134=180
Subtract 134 from both sides.
2x=180-1342x=180−134
2x=462x=46
Divide both sides by 2.
x=\frac{46}{2}x=
2
46
x=23}