Question

Please Prove question no. 21 with all the steps.

Answer 1

Answer:

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 !

Answer 2

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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