Question 77

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

Solution

$$ \angle ABD $$ = $$ \angle MBD $$=?(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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