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Question

Internal bisector of A of triangle ABC meets side BC at D. A line drawn through D perpendicular to AD intersects the side AC at E and the side AB at F. If a, b, c represent sides of ΔABC then

A
AE is HM of b and c
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B
AD =2bcb+ccosA2
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C
EF=4bcb+csinA2
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D
the triangle AEF is isosceles
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Solution

The correct options are
A AE is HM of b and c
B EF=4bcb+csinA2
C the triangle AEF is isosceles
D AD =2bcb+ccosA2
We have ΔABC=ΔABD+ΔACD

12 bc sinA=12cADsinA2+12b×ADsinA2 AD =2bcb+ccosA2

Again AE = AD secA2

=2bcb+c AE is HM of b and c.
EF =ED+DF=2DE=2×AD

tanA2=2×2bcb+c×cosA2×tanA2

=4bcb+csinA2 As AD EF and DE=DF and AD
is bisector AEF is isosceles.
Hence A, B, C and D are correct answers.

368939_42679_ans.jpg

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