Question

$Y$ is any point inside a triangle $ABC$. The bisector of $\angle AYB$, $\angle BYC$ and $\angle CYA$ meet the sides, $AB,BC$ and $CA$ in point $D,E$ and $F$ respectively.
Then, $AD\times BE$ = $DB\times FA$.

• A. True
• B. False

Answer 1

The correct option is B False

Given: $△ABC$, CX is bisector of $\angle C$,
$\angle ACY=\angle BCX$ ...(I)
and $AX=AY$
In $△AXY$,
$\angle AXY=\angle AYX$ (Angles opposite to equal sides)...(II)
Now, $\angle XYC=\angle AXB=180$ (Angles on a straight line)
$\angle AYX+\angle AYC=\angle AXY+\angle BXY$
hence, $\angle AYC=\angle BXY$....(III)
Also, In $△AYC$ and $△BXC$
$\angle AYC+\angle YCA+\angle CAY=\angle BXC+\angle BCX+\angle XBC=180$ (Angles sum property)
$\angle CAY=\angle XBC$ (From I and III)
or $\angle CAY=\angle ABC$