RHS Criteria of congruence of triangles
Suppose, you have taken a rectangular sheet in your hands.
You cut it along its diagonal.
Thus, this rectangular sheet is divided in two triangles.
Here, it is important to know that any such two triangles would always be congruent to each other.
This can be proved practically by superposition.
Or we can use other method of RHS criterion to prove this congruency.
Let’s understand the RHS criterion for triangles.
Suppose two triangles are given as
$AB$
is perpendicular to
$BC$
and
$PQ$
is perpendicular to
$QR$
, so both triangles are right angled triangles
Suppose we are given
$AB=PQ$
,
$AC=PR$
and
$∠ABC=∠PQR$
i.e.
$90°$
$RHS$
criteria specifies right angle, hypotenuse and side
RHS congruence criterion specifies that if hypotenuse and one side of two right angled triangles are equal, then both triangles are congruent.
By
$RHS$
congruence rule both the triangles are congruent
Let’s understand RHS congruence rule with another example.
Suppose we are given with figure as such and AC = BD
$∠ABC$
=
$∠DCB$
=
$90°$
in
$△ABC$
and
$△BCD$
.
Hypotenuse of both triangles are equal and side
$BC$
is common in both triangle
So both the triangles
$BCD$
&
$CBA$
are congruent by
$RHS$
congruence.
Revision
RHS criteria for congruence specifies that hypotenuse and one side of two right angle triangles are equal, both triangles are congruent.
The End