5. In the given figure, AB = DC and angleABC = angleDCB. Prove that
(a) ∆ABC = ∆DCB
(b) angleA = angleD
(c) ∆AOB = ∆DOC
(d) ∆OBC is isosceles
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In the given figure, AB = DC and ∠BC = ∠DCB. Prove that
(a) ∆ABC = ∆DCB
(b) ∠A = ∠D
(c) ∆AQB = ∆DQC
(d) ∆QBC is isosceles
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★
∆ABC & ∆DCB :-
- AB = DC
- ∠ABC = ∠DCB
- BC = same base
Therefore proved ∆ABC = ∆DCB [ SAS criterion ]
★ ∠A = ∠D
Since BC is the same base for both the angles and according to the theorem angles on the same base is equal
★ Proof :-
- ∠AQB = ∠DQC
- AB = DC
- ∠BAC = ∠BDC
∆AQB = ∆DQC
BQ = QC [ CPCTC ]
★ As, BQ = QC [ proved above ]
So, ∆OBC is an isosceles triangle
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