The side A C of a triangle A B C is produced to point E so that CE - 1 - 2 AC- D is the mid-point point of B C and ED produced meets AB at F and CP and DQ are parallel to AB - then FD - 1 - 4 FE -

The correct option is B False #160; Here D is the midpoint of BC and DP is parallel to AB #160; therefore P is the midpoint of AC and PD=AB/2 ...

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Byju's Answer

Standard IX

Mathematics

Equal Intercept Theorem

The side A ...

Question

The side $A C$ of a triangle $A B C$ is produced to point $E$ so that $C E = \frac{1}{2}$ $A C$ . $D$ is the mid-point point of $B C$ and $E D$ produced meets $A B$ at F and CP and DQ are parallel to AB . then $F D = \frac{1}{4} F E$ ?

True

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False

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Solution

The correct option is B False
$H e r e D is the midpoint of BC and DP is parallel to AB$
$therefore P is the midpoint of AC and PD= \frac{A B}{2}$
$F r o m triangle AEF$
$A E | | P D | | C R a n d A P = \frac{A E}{3}$
$H e n c e D F = \frac{E F}{3}$
$H e n c e t h e g i v e n s t a t e m e n t i s f a l s e$

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The side AC of a triangle ABC is produced to point E so that CE=12 AC. D is the mid-point point of BC and ED produced meets AB at F and CP and DQ are parallel to AB . then FD=14FE ?

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The side $A C$ of a triangle $A B C$ is produced to point $E$ so that $C E = \frac{1}{2}$ $A C$ . $D$ is the mid-point point of $B C$ and $E D$ produced meets $A B$ at F and CP and DQ are parallel to AB . then $F D = \frac{1}{4} F E$ ?