A problem by Yahia El Haw

Level pending

$ABCD$ is a parallelogram. $E$ is a point on it. $\overrightarrow { ED }$ is a bisector for $\hat { (ADC) }$ . $\overrightarrow { EC }$ is a bisector for $\hat { (DCB) }$ . $F$ is the midpoint of $DC$ .

Choose the correct answer

EF=FC

EF<FC

EF>FC

1 solution

Yahia El Haw
Nov 19, 2016

$\because$ $AD$ // $BC$

$\therefore$ $\hat { (ADC) }$ + $\hat { (DCB) }$ = ${ 180 }^{ \circ }$

$\therefore$ $\frac { 1 }{ 2 }$ $\hat { (ADC) }$ + $\frac { 1 }{ 2 }$ $\hat { (DCB) }$ = ${ 90 }^{ \circ }$

$\therefore$ $\hat { (EDC) }$ + $\hat { (DCE) }$ = ${ 90 }^{ \circ }$

$\because$ The Measurements of $\triangle$ = ${ 180 }^{ \circ }$

$\therefore$ $\hat { (DEC) }$ = ${ 90 }^{ \circ }$

$\because$ $DF$ = $FC$

$\therefore$ $EF$ = $\frac { 1 }{ 2 }$ $DC$

$\therefore$ $EF$ = $FC$

A problem by Yahia El Haw

1 solution

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