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Byju's Answer

Standard XII

Mathematics

Distance Formula

AD is the med...

Question

$A D$ 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 $A B$ at $E$ and $A C$ at $F$ . Prove that $E F ∥ B C$ .

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Solution

In $Δ D A E, D E$ bisect angle $A D B$

So we have

$\frac{D A}{D B} = \frac{A E}{E B}$ ... (1)

Similarly in $Δ D A C, D E$ bisect angle $A D C$

we get

$\frac{D A}{D C} = \frac{A F}{A C}$

$(D C = D B)$

$\frac{D A}{D B} = \frac{A F}{F C}$ ... (2)

from eq (1) and eq (2)

$\frac{A E}{E B} = \frac{A F}{F C}$

ln $Δ A B C$

$E F | | B E (. . . B P T)$

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AD is the median of △ABC and bisectors of ∠ADB and∠ADC are DE and DF, which meet AB at E and AC at F. Prove that EF∥BC.

In ΔDAE,DE bisect angle ADBSo we have DADB=AEEB ... (1)Similarly in ΔDAC,DE bisect angle ADC we getDADC=AFAC(DC=DB)DADB=AFFC ... (2)from eq (1) and eq (2)AEEB=AFFCln ΔABCEF||BE(...BPT)

$A D$ 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 $A B$ at $E$ and $A C$ at $F$ . Prove that $E F ∥ B C$ .