Question

The perpendicular bisector of side ab of ∆abc intersects the extension of side ac at
d. find the measure of ∠abc if m∠cbd = 16° and m∠acb = 118°.

Answer 1

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°.

Answer 2

$\rm \angle ABC = 23^ \circ$

Step-by-step explanation:

Given :

$\rm \angle CBD = 16 ^ \circ$

$\rm \angle ACB = 118 ^ \circ$

Solution :

According to SAS postulate,

$\rm \bigtriangleup ADM = \bigtriangleup BDM$

$\rm \angle MAD = \angle MBD \;\;\;\;\;\;(Corresponding \;Parts \;of \;Congruent \;Triangles\; are \; Congruent)$Now,

$\rm \angle ABC + \angle ACB +\angle BAC=180^\circ$

$\rm x^\circ + 118^\circ +( x^\circ +16^\circ) = 180^\circ$        (refer the image)

$\rm 2x = 46^\circ\\\rm x = 23^\circ$

Therefore,

$\rm \angle ABC = 23^ \circ$

For more information, refer to the link given below

https://brainly.com/question/10376207?referrer=searchResults