Question

In the following figure, $\angle BAC={60}^o$ and $\angle ABC={65}^o$.
Prove that :
$DC>DF$

Answer 1

Given $\angle BAC=60$

$\angle ABC={65}^o$

know , sum of angles on straight line $={180}^o$

$\therefore \angle AEC+\angle BEC={180}^o$

$90+\angle BEC={180}^o$

$\angle BEC={90}^o$

$\therefore$ In $△BEC$

$\angle EBC+\angle BEC+\angle ECB={180}^o$

${65}^o+90+\angle FCB={180}^o$

$\angle ECB=180-{155}^o$

${25}^o$

$\therefore$ In $△DFC$

$\angle DFC=180-90-25$

$={65}^o$

We know,

side opposite to greater angle is greatest

fin a,$\angle DFC>\angle DCF$

$\therefore DC>DF$.