Mathematics
In the following figure, OA = OC and AB = BC. Prove that :
(i) ∠AOB = 90°
(ii) △ AOD ≅ △ COD
(iii) AD = CD
Triangles
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Answer
(i) In △ AOB and △ COB,
⇒ OA = OC (Given)
⇒ AB = BC (Given)
⇒ OB = OB (Common side)
∴ △ AOB ≅ △ COB (By S.S.S. axiom).
We know that,
Corresponding parts of congruent triangles are equal.
∴ ∠AOB = ∠COB = x (let)
From figure,
AC is a straight line.
∴ ∠AOB + ∠COB = 180°
⇒ x + x = 180°
⇒ 2x = 180°
⇒ x = = 90°.
∴ ∠AOB = 90°.
Hence, proved that ∠AOB = 90°.
(ii) We know that,
Vertically opposite angles are equal.
∴ ∠AOD = ∠COB = 90° and ∠COD = ∠AOB = 90°.
In △ AOD and △ COD,
⇒ OA = OC (Given)
⇒ ∠AOD = ∠COD (Both equal to 90°)
⇒ OD = OD (Common side)
∴ △ AOD ≅ △ COD (By S.A.S. axiom).
Hence, proved that △ AOD ≅ △ COD.
(iii) Since, △ AOD ≅ △ COD.
We know that,
Corresponding parts of congruent triangles are equal.
∴ AD = CD
Hence, proved that AD = CD.
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