AD is perpendicular to the internal bisector of$$ \angle ABC $$ of $$ \triangle$$ ABC. DE is drawn through D and parallel to BC to meet AC at E. If the length of AC is 12 cm, then the length of AE (in cm)is
$$ \angle ABD $$ = $$ \angle MBD $$=?(angle bisector)
BD ⊥AM
$$ \angle BDA $$= $$ \angle BDM $$=90°
It happen only in equilateral and isosceles triangle
AD=DM
i.e.AD=AM/2
Given DE || BC
From Thales theorem
E will be mid point of AC.
AC=12cm.
So,
AE=6cm
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