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Example 20: In the adjoining figure, AB = AC and D is the mid-point of BC.

(a) State the three pairs of equal parts in DADB and DADC.
(b) Is DADB ≅ DADC? Give reasons.
(c) Is ∠BAD = ∠CAD? Why?

Solution:

(a) AB = AC (Given),
AD = AD (Common side)

BD = CD (D is the midpoint of BC)
(b) From part (a) above, DADB ≅ DADC (By SSS congruence criterion)

(c) Yes, ∠BAD = ∠CAD (C.P.C.T.)
Example 21: Use ASA congruence rule and verify that DAOC ≅ DBOD?

D

A 70° 4
O
30°
4 B

70°

C
Solution: In the two triangles, AOC and BOD, ∠C = ∠D (Each 70°)

Also, ∠AOC = ∠BOD = 30° (Vertically opposite angles)

So, ∠A of DAOC = 180° – (70° + 30°) = 80° (Using angle sum property of a triangle)

Similarly, ∠B of DBOD = 180° – (70° + 30°) = 80°
Thus, we have ∠A = ∠B, AC = BD and ∠C = ∠D

Now, side AC is between ∠A and ∠C and side BD is between ∠B and ∠D.

So, by ASA congruence rule, DAOC ≅ DBOD.

Example 22: In the adjoining figure, DA ⊥ AB, CB ⊥ AB and AC = BD. D C
State the three pairs of equal parts in DABC and DBAD.

Also, show that DABC ≅ DBAD A B

Solution: The three pairs of equal parts are:
∠ABC = ∠BAD (Each 90°)

AC = BD (Given)

AB = BA (Common side)
From the above,

DABC ≅ DBAD (By RHS congruence rule).

189 The Triangle and Its Properties