JEE Advanced · 2006 · 09 Apr 2006 · Properties of Triangle

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

△ A B C

then

(A)AE is HM of b and c✓ Correct
(B) $A D = \frac{2 b c}{b + c} cos \frac{A}{2}$ ✓ Correct
(C) $E F = \frac{4 b c}{b + c} sin \frac{A}{2}$ ✓ Correct
(D)the triangle AEF is isosceles✓ Correct

Answer: (A), (B), (C), (D)

Solution

Using the angle bisector property and area considerations:

Area of

△ A B C =

Area of

△ A B D +

Area of

△ A C D

\frac{1}{2} b c sin A = \frac{1}{2} c \cdot A D sin \frac{A}{2} + \frac{1}{2} b \cdot A D sin \frac{A}{2}

b c sin A = A D (b + c) sin \frac{A}{2}

Using

sin A = 2 sin \frac{A}{2} cos \frac{A}{2}

2 b c sin \frac{A}{2} cos \frac{A}{2} = A D (b + c) sin \frac{A}{2}

A D = \frac{2 b c}{b + c} cos \frac{A}{2}

Since

E F ⊥ A D

A E = A D sec \frac{A}{2} = \frac{2 b c}{b + c}

This is the harmonic mean of

b

and

c

.

For

E F

: Since

D

is the midpoint of

E F

(by symmetry about the angle bisector):

E F = 2 D E = 2 \cdot A D tan \frac{A}{2} = \frac{4 b c}{b + c} sin \frac{A}{2}

Since

A D

is the perpendicular bisector of

E F

and also the angle bisector of

∠ A

, triangle

A E F

is isosceles with

A E = A F

.

All four options are correct.

JEE Advanced · 2006 · 09 Apr 2006 · One or more correct options

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JEE Advanced · 2006 · 09 Apr 2006 · Properties of Triangle

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