Question

State whether the following statement is true or false.

$D$ and $E$ are respectively the points on equal sides $AB$ and $AC$ of an isosceles triangle $ABC$ such that $B,C,E$ and $D$ are concyclic, if $O$ is point of intersection of $CD$ and $BE$, then $AO$ is the bisector of line segment $DE$

• A. True
• B. False

Answer 1

The correct option is A True

Since points $B,C,D$ and $E$ are Concyclic.

Sum of opposite angles of cyclic quadrilateral is ${180}^{\circ }$

So,$\angle 1+\angle 3={180}^{\circ }$

and $\angle 2+\angle 4={180}^{\circ }$

Also, $△ABC$ is the Isosceles triangle.

$AB=AC$(Given)

$\angle 3=\angle 4$ since when opposite sides are equal angle opposite to them are equal.

Given:$\angle 1+\angle 4={180}^{\circ }$

and $\angle 2+\angle 3={180}^{\circ }$

But these are interior angles on the same side of transversal $BD$ and $EC$.So if the sum of interior angles on the same side of transversal is ${180}^{\circ }$, then lines are parallel.

$\because DE\parallel BC$

$\angle 4=\angle 5$

and $\angle 3=\angle 6$

when lines are parallel, corresponding angles are equal.

But,$\angle 3=\angle 4$

$\Rightarrow \angle 5=\angle 6$

$\therefore AD=AE$ since If opposite angles in a triangle are equal, the side opposite to them are equal.

From the figure $AM\perp DE$

In $△AMD$ and $△AME$

$AD=AE$ proved above.

$\angle AMD=\angle AME$ each being ${90}^{\circ }$

$AM$ is common.

$△AMD\cong △AME$ by RHS congruency rule

$\Rightarrow DM=ME$ by C.P.C.T

From the fig.in the question,$AO$ which passes through $M$, as $\angle AME+\angle OME={180}^{\circ }$,showing points $A,M$ and $O$ are collinear.

which shows, Segment $AO$ is the bisector of line segment $DE\because DM=ME$.