Question

In a \[\Delta ABC\], the bisector of \[\angle A\] passes through its circumcenter. Prove that \[AB=BC\].

Answer 1

Hint: Try to divide the given triangle into different triangles containing \[AB\] and \[AC\] and use similarity of triangles.

Complete step-by-step answer:
Given a \[\Delta ABC\] in which the bisector of \[\angle A\] passes through its circumcenter.
We have to prove that \[AB=AC\].

In above figure, \[\Delta ABC\] has a circumference with circumcenter \[G\] and \[AD\] is angle bisector of \[\angle A\].
Since, \[AD\] is an angle bisector, therefore \[\angle BAD=\angle DAC....\left( i \right)\]
Since, \[G\] is circumcenter,
Therefore, \[GA=GB=GC=R=\text{circumradius of circle }....\left( ii \right)\]
In \[\Delta GAB\], since \[GA=GB\][by equation \[\left( ii \right)\]]
We know that angles opposite to equal sides are equal.
Therefore, \[\angle GAB=\angle GBA....\left( iii \right)\]
Also in \[\Delta GAC\], since \[GA=GC\][By equation \[\left( ii \right)\]]
We know that angles opposite to equal sides are equal.
Therefore, \[\angle GAC=\angle GCA....\left( iv \right)\]
From equation\[\left( i \right)\],\[\left( iii \right)\]and \[\left( iv \right)\]
We get \[\angle GAB=\angle GBA=\angle GAC=\angle GCA....\left( v \right)\]
Now, in \[\Delta GAC\] and \[\Delta GAB\],
\[GA=GA\left[ \text{common} \right]\]
\[\angle GAC=\angle GAB\left[ \text{By equation }\left( i \right) \right]\]
\[\angle GBA=\angle GCA\left[ \text{By equation }\left( v \right) \right]\]
Therefore, \[\Delta GAC\] and \[\Delta GAB\] are congruent by AAS [Angle-Angle-Side] property.
We know that corresponding parts of congruent triangles are equal.
Therefore, we get, \[AB=AC\]
Hence proved.

Note: Students often mistake AAA as the criteria for congruence which is only the criteria for similarity of triangles. Mistakes could be committed in writing corresponding parts of two triangles.