Repeat Exercise $7$, but use the $SAS$ method.

A triangle is given.

A triangle congruent to $\Delta ACU$ using Side-Angle-Side postulate,

Consider the following ways-

Draw a line $L$.

Choose a point $Q$on line $L$.

Set compass on radius of $CU$.

Take point $Q$as center and draw an arc on line $L$. Mark that point as $R$.

Draw an arc by taking $C$as center which touches the sides $CA$ and $CU$.

With same radius draw an arc with $Q$as center.

Then take the distance between the intersection points $D$ and $E$.

Draw that distance by keeping $S$as center and draw an arc.

Draw a ray through the intersecting point and mark it as $T$.

Take $CA$as length draw an arc with $Q$as center on ray $QT$.

Mark that intersection point as $P$.

Draw a segment joining $P$and $R$.

Side-Angle-Side theorem states that,

if two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle,

then these two triangles are congruent.

So, $\Delta ACU$is congruent to $\Delta PQR$.

Since, $\angle CU\cong \angle QR,\angle C\cong \angle Q$and $\angle AC\cong \angle PQ$then $\Delta ACU\cong \Delta PQR$.

Therefore, by $SAS$method role="math" $\Delta ACU\cong \Delta PQR$.