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Answer :
In \(\triangle{ABC}\) and \(\triangle{ACD}\), we have,
DB = DC ...(given)
\(\angle{ADB}\) = \(\angle{ADC}\) ...(Since, AD is the perpendicular bisector of BC)
and AD is the Common side.
Therefore, \(\triangle{ABD}\) \(\displaystyle \cong \) \(\triangle{ACD}\) ...(By SAS congruency test)
Therefore, AB = AC ...(By CPCT)
Hence, \(\triangle{ABC}\) is an isosceles triangle.
Hence, proved.