Question

In $\Delta XYZ$, $XM$ is the bisector of $\angle x$ and $AB$ is perpendicular to $XM,$ cutting it at $D$. Then which of the following relations holds true?

• A. $XD=DM$
• B. $YM=MZ$
• C. $XA=XB$
• D. $DM=BZ$

Answer 1

The correct option is C $XA=XB$

Given $\Delta XYZ,XM$ is a bisector of $\angle X$ and $AB$ is perpendicular to $XM$, cutting at $D$.

In $\Delta AXD$ and $\Delta DXB$

$\angle AXD=\angle DXB$

$\angle XDB=\angle XDA={90}^0$

Side XD is common

$\therefore \Delta AXD\cong \Delta DXB$

Then all sides of $\Delta AXD$ and $\Delta DXB$ are congruent

Then $XA=XB$ and $AD=DB$

Hence, option C is correct.