91Ó°ÊÓ

The labelled diagram satisfying the given statement is:

The statement is: If$∠A$ and$∠B$are the base angles of isosceles $â–³ABC$, and the bisector of$∠A$ meets$\stackrel{¯}{BC}$ at$X$and the bisector of$∠B$ meets$\stackrel{¯}{AC}$ at $Y$, then $\stackrel{¯}{AX}â‰\dots \stackrel{¯}{BY}$.

Consider the isosceles triangle be $â–³ABC$.

Consider the two equal sides be$AC$ and $AC$.

Therefore, it is given that $\stackrel{¯}{CA}â‰\dots \stackrel{¯}{CB}$.

As,$∠A$ and$∠B$ are the base angles of isosceles $△ABC$, and the bisector of$∠A$ meets$\stackrel{¯}{BC}$ at$X$and the bisector of$∠B$ meets$AC$at $Y$.

Therefore, $∠CAX=∠BAX$and $∠CBY=∠ABY$.

Therefore, it is given that $∠CAX=∠BAX$and $∠CBY=∠ABY$.

It is to be proved that $\stackrel{¯}{AX}â‰\dots \stackrel{¯}{BY}$.

By using the angle addition postulate it can be noticed that:

$∠CAB=∠CAX+∠BAX$

As,$∠CAX=∠BAX$,therefore, it can be noticed that:

$$\begin{gathered} ∠CAB=∠CAX+∠BAX \\ =∠CAX+∠CAX \\ =2∠CAX \end{gathered}$$

By using the angle addition postulate it can be noticed that:

$∠CBA=∠CBY+∠ABY$

As,$∠CBY=∠ABY$,therefore, it can be noticed that:

$$\begin{gathered} ∠CBA=∠CBY+∠ABY \\ =∠CBY+∠CBY \\ =2∠CBY \end{gathered}$$

As, $\stackrel{¯}{CA}â‰\dots \stackrel{¯}{CB}$,therefore, by using the theorem 4-2, it can be noticed that $∠CAB=∠CBA$.

As, $∠CAB=∠CBA$, therefore, it can be noticed that:

$\begin{array}{c}∠CAB=∠CBA\\ 2∠CAX=2∠CBY\\ ∠CAX=∠CBY\end{array}$

Therefore, $∠CAXâ‰\dots ∠CBY$.

In the triangles$△CXA$ and $△CYB$, it can be noticed that the angle$∠C$ is common.

Therefore, $∠Câ‰\dots ∠C$by using the reflexive property.

Therefore, in the triangles $â–³CXA$and $â–³CYB$, it can be noticed that $∠Câ‰\dots ∠C,\stackrel{¯}{CA}â‰\dots \stackrel{¯}{CB}$and $∠CAXâ‰\dots ∠CBY$.

Therefore, the triangles$â–³CXA$ and$â–³CYB$ are the congruent angles by using the ASA postulate.

The triangles$â–³CXA$ and$â–³CYB$ are the congruent triangles.

Therefore, by using the corresponding parts of congruent triangles it can be said that $\stackrel{¯}{AX}â‰\dots \stackrel{¯}{BY}$.

The proof in two-column form is: