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kumpel [21]
3 years ago
12

The perpendicular bisector of side ab of ∆abc intersects the extension of side ac at

Mathematics
1 answer:
Triss [41]3 years ago
6 0

Answer:

\angle ABC=23^{\circ}

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°.

\angle ABD=\angle ABC+\angle CBD

\angle ABD=x+16

In triangle ABD,  DM is perpendicular bisector of AB.

In triangle ADM and BDM,

AM\cong BM              (Definition of perpendicular bisector)

\angle AMD\cong \angle BMD            (Definition of perpendicular bisector)

DM\cong DM              (Reflection property)

By SAS postulate,

\triangle ADM\cong \triangle BDM

\angle MAD\cong \triangle MBD            (CPCTC)

\angle MAD=x+16

\angle 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

x+118+(x+16)=180

2x+134=180

Subtract 134 from both sides.

2x=180-134

2x=46

Divide both sides by 2.

x=\frac{46}{2}

x=23}

Therefore, the measure of ∠ABC is 23°.

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