Premium Online Home Tutors
3 Tutor System
Starting just at 265/hour
Answer :
In \(\triangle{ABC}\), we have
AB = AC ...(Given)
Also, \(\angle{B}\) = \(\angle{C}\) ...(Since corresponding angles of equal sides are equal)
Multipling both sides by \(\frac{1}{2} \) , we get,
(\(\frac{1}{2} \) ) \(\angle{B}\) = (\(\frac{1}{2} \) ) \(\angle{C}\)
\(\therefore \) \(\angle{OBC}\) = \(\angle{OCB}\)
Also, it is given that, OB and OC are bisectors of \(\angle{B}\) and \(\angle{C}\), respectively,
Therefore, \(\angle{OBA}\) and \(\angle{OCA}\)
Therefore, OB = OC
...(\(\because \) corresponding sides of equal angles are equal)
Hence, part (i) is proved.
In \(\triangle{ABO}\) and \(\triangle{ACD}\), we have,
AB = AC ...(Given)
\(\angle{OBA}\) = \(\angle{OCA}\) ...(from part(i))
OB = OC ...(proved earlier)
Therefore, \(\triangle{ABO}\) \(\displaystyle \cong \) \(\triangle{ACO}\) ...(By SAS congruency test)
Thus, \(\angle{BAO}\) = \(\angle{CAO}\) ...(By CPCT)
Hence, AO bisects \(\angle{A}\) is proved.