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Question
In the following figure, OA = OC and AB = BC.
Prove that:
(i) ∠AOB = 90o
(ii) ΔAOD ≅ ΔCOD
(iii) AD = CD
Solution
Given:
In the figure, OA=OC, AB =BC
We need to prove that,
AOB = 90°
(i) In ΔABO and ΔCBO,
AB = BC ...[given ]
AO = CO ...[ given ]
OB = OB ...[ common ]
∴By Side-Side-Side criterion of congruence, we have
ΔABO ≅ ΔCBO
The corresponding parts of the congruent triangles are congruent.
∴∠ABO = ∠CBO ...[c. p.c.t. ]
⇒ ∠ABD = ∠CBD
and ∠AOB = ∠COB ...[c. p.c t ]
We have
∠AOB + ∠COB = 180° .....[ linear pair ]
⇒ ∠AOB = ∠ COB= 90° and AC ⊥ BD
(ii) In ΔAOD and ΔCOD,
OD = OD ...[ common ]
∠AOD = ∠COD ...[ each=90° ]
AO = CO ...[ given]
∴By Side-Angel-Side criterion of congruence, we have
ΔAOD ≅ ΔCOD
(iii) The corresponding parts of the congruent
triangles are congruent.
∴AD = CD ...[c. p.c t ]
Hence proved.
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