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Congruence of Simple Geometrical Figures

Two plane figures, say, P and P are said to be congruent, if the trace-copy of P fits exactly on
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that of P . We write this as P ≅ P .
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Congruence among Line segments: If two line segments have the same (i.e. equal) length, they
are congruent. Also, if two line segments are congruent, they have the same length.
Suppose two line segments AB and CD are congruent.
A B
Then, we write this as AB CD≅ . C D
However, it is common to write it as AB CD= .
C R
Congruence of Angles: Two angles, say ∠ABC and ∠PQR, are
congruent, if their measures are equal. We write this as ∠ABC ≅ ∠PQR B Q
or as m ∠ABC = m∠PQR or simply as ∠ABC = ∠PQR. A P

Congruence of Triangles: Two triangles are congruent if they are copies of each other and when
superimposed, they cover each other exactly.
(a) A (b) P

B C Q R
ΔABC and ΔPQR have the same size and shape. They are congruent.
So we would express this as ΔABC ≅ ΔPQR.

This means that, when you place ΔPQR on ΔABC , P falls on A, Q Remember
falls on B and R falls on C, also PQ falls along AB,QR falls along • Two circles are congruent, if
they have equal radii.
BC and PR falls along AC . • Two squares are congruent if

If under a given correspondence, two triangles are congruent, they have equal sides.
then their corresponding parts (i.e., angles and sides) that match • Two rectangles are congruent if
with each other are equal. they have equal length and breadth.

This shows that while talking about congruence of triangles, not
only the measures of angles and lengths of sides matter, but also the matching of vertices.
In the above case, the correspondence is A ↔ P, B ↔ Q, C ↔ R

We may write this as ABC ↔ PQR
Example 17: In the given figure, if DPQR ≅ DYXZ under the correspondence
PQR ↔ YXZ, write all the corresponding congruent parts of the triangles.

Solution: PQ ↔ YX, QR ↔ XZ, RP ↔ ZY, ∠P = ∠Y, ∠Q = ∠X, ∠R = ∠Z
Example 18: Without drawing the triangles, state the correspondence
between the sides and the angles of the following pairs of triangles.
(a) DABC ≅ DPQR (b) DDEF ≅ DBCA

Solution:
(a) A ↔ P, B ↔ Q, C ↔ R (b) D ↔ B, E ↔ C, F ↔ A
AB = PQ, BC = QR, AC = PR DE = BC, EF = CA, DF = BA

∠A = ∠P, ∠B = ∠Q, ∠C = ∠R ∠D = ∠B, ∠E = ∠C, ∠F = ∠A

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