Please Prove question no. 21 with all the steps.
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Proof below ↓↓↓
Step-by-step explanation:
In Δ ABD and Δ ADE ,
AD = AE [ Given ]
So ∠ADE = ∠AED [ base ∠s of isosceles Δ ]
⇒ 180° - ∠DE = 180° - ∠AED
⇒ ∠ADB = ∠AEC [ From figure ]
BD = EC [ Given ]
Hence Δ ABD ≅ Δ ACE [ S.A.S ]
⇒ AB = AC [ c.p.c.t ]
Hence Proved !
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Answered by
1
In Δ ABD and Δ ADE ,
AD = AE [ Given ]
So ∠ADE = ∠AED [ base ∠s of isosceles Δ ]
⇒ 180° - ∠DE = 180° - ∠AED
⇒ ∠ADB = ∠AEC [ From figure ]
BD = EC [ Given ]
Hence Δ ABD ≅ Δ ACE [ S.A.S ]
⇒ AB = AC [ c.p.c.t ]
Hence Proved !
HOPE IT HELPS U ✌️✌️
AD = AE [ Given ]
So ∠ADE = ∠AED [ base ∠s of isosceles Δ ]
⇒ 180° - ∠DE = 180° - ∠AED
⇒ ∠ADB = ∠AEC [ From figure ]
BD = EC [ Given ]
Hence Δ ABD ≅ Δ ACE [ S.A.S ]
⇒ AB = AC [ c.p.c.t ]
Hence Proved !
HOPE IT HELPS U ✌️✌️
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