Answer:
Solution given;
<BCA=<ACD
<ABC=<ADC
now
In traingle ABC & ∆ ACD
AC=AC[common base]
<BCA=<ACD[bisector bisect it]
<ABC=<ADC[Given]
∆ABC is congruent to ∆ ACD by S.A.A. axiom
AB is congruent to AD.
[ corresponding sides of a congruent triangle are equal]
Hence proved.
Opposite sides of ∠BAC and ∠CAD is equal BC = DC .
Therefore, AB is congruent to AD.
Given that,
AC bisects angle BCD ,
Angle ABC is congruent to angle ADC
We have to prove,
AB is congruent to AD.
According to the question,
AC bisects ∠BCD,
Angle ABC is congruent to angle ADC
Then, In triangle ABC & ∆ ACD
AC =AC [common base in the triangle]
<BCA = <ACD [bisector bisect it]
<ABC = <ADC [Given]
If one angle between two adjacent sides of a triangle is similar to angle between two sides of another triangle then two triangles are congruent is called SAS-Congruence-Axiom.
Then,
By SAS-Congruence-Axiom
∆ABC is congruent to ∆ACD
AB is congruent to AD. [corresponding sides of a congruent triangle are equal]
Then,
BC (opposite side of ∠BAC) = DC (opposite side of ∠CAD),
BC = DC
Opposite sides of ∠BAC and ∠CAD is equal BC = DC .
Therefore, AB is congruent to AD.
Hence proved.
To know about Triangles click the link given below.
https://brainly.com/question/668486
