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Lines in a Triangle

The median AD...

Question

The median AD of $Δ A B C$ meets BC at D. The internal bisectors of $∠ A D B$ and $∠ A D C$ meet $A B$ and $A C$ at $E$ and $F$ respectively. Then $E F$ :

is perpendicular to

A D

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is parallel to

B C

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divides

A D

in the ratio of

A B : A C

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None of these

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Solution

The correct option is B is parallel to $B C$
In $△ D A E, D E$ bisect $∠ A D B$
$\Rightarrow \frac{D A}{D B} = \frac{A E}{E B}$ ...... $(1)$
Similarly, in $△ D A C$ ,
$D E$ bisect $∠ A D C$
$\frac{D A}{D C} = \frac{A F}{F C}$
$⟹ D C = D B$
$∴ \frac{D A}{D B} = \frac{A F}{F C}$ ..... $(2)$
$\frac{A E}{E B} = \frac{A F}{F C}$
In $△ A B C, E F ∥ B C$

Suggest Corrections

Similar questions

A D

is a median of

△ A B C

. The bisector of

∠ A D B

and

∠ A D C

meet

A B

and

A C

E

and

F

respectively. Prove that

E F | | B C

A D

is the median of

△ A B C

and bisectors of

∠ A D B

and

∠ A D C

are

D E

and

D F

, which meet

A B

E

and

A C

F

. Prove that

E F ∥ B C

A D

is internal angle bisector of

△ A B C

∠ A

and

D E

perpendicular to

A D

which intersects

A C

E

and meets

A B

F

, then:

Q. 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

Δ A B C

then

Question 28

In $Δ A B C$ , AD is the bisector of $∠ A$ meeting BC at D, $C F ⊥ A B$ and E is the mid-point of AC. Then, median of the triangle is

a) AD
b) BE
c) FC
d) DE

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The median AD of ΔABC meets BC at D. The internal bisectors of ∠ADB and ∠ADC meet AB and AC at E and F respectively. Then EF:

The correct option is B is parallel to BCIn △DAE,DE bisect ∠ADB⇒DADB=AEEB ...... (1)Similarly, in △DAC,DE bisect ∠ADCDADC=AFFC⟹DC=DB∴DADB=AFFC ..... (2)AEEB=AFFCIn △ABC,EF∥BC

The median AD of $Δ A B C$ meets BC at D. The internal bisectors of $∠ A D B$ and $∠ A D C$ meet $A B$ and $A C$ at $E$ and $F$ respectively. Then $E F$ :