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प्रश्न
In the given figure, AC = AE. Show that:
- CP = EP
- BP = DP
उत्तर
In ΔADC and ΔABE,
∠ACD = ∠AEB ...(Angles in the same segment)
AC = AE ...(Given)
∠A = ∠A ...(Common)
∴ ΔADC ≅ ΔABE ...(ASA postulate)
`=>` AB = AD
But AC = AE
∴ AC – AB = AE – AD
`=>` BC = DE
In ΔBPC and ΔDPE
∠C = ∠E ...(Angles in the same segment)
BC = DE
∠CBP = ∠CDE ...(Angles in the same segment)
∴ ΔBPC ≅ ΔDPE ...(ASA Postulate)
`=>` BP = DP and CP = PE ...(C.P.C.T.)
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