In $\Delta ABC$,$AB = 8 \, cm$ and $BC = 5 \, cm$,then $AC > \ldots \ldots \ldots cm$.

It is given that $\triangle ABC \cong \triangle RPQ$. Is it true to say that $BC = QR$? Why?

In the given figure,$BA \perp AC$ and $DE \perp DF$ such that $BA = DE$ and $BF = EC$. Show that $\triangle ABC \cong \triangle DEF$.

Show that in a quadrilateral $ABCD$,$AB + BC + CD + DA < 2(BD + AC)$.

$ABC$ is an isosceles triangle with $AB = AC$ and $D$ is a point on $BC$ such that $AD \perp BC$. To prove that $\angle BAD = \angle CAD$,a student proceeded as follows:
In $\triangle ABD$ and $\triangle ACD$:
$AB = AC$ (Given)
$\angle B = \angle C$ (because $AB = AC$)
and $\angle ADB = \angle ADC$
Therefore,$\triangle ABD \cong \triangle ACD$ $(AAS)$
So,$\angle BAD = \angle CAD$ $(CPCT)$
What is the defect in the above arguments?

In $\Delta ABC$ and $\Delta PQR$,if $\frac{AB}{PQ} = \frac{BC}{RQ} = \frac{AC}{PR} = 1$,then $\Delta ABC \cong \Delta \ldots \ldots \ldots$