Question

( 1 ) In Fig AC = AE, AB = AD and angle BAD = angle EAC. prove that BC = DE.

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Answer 1

Answer:

We are given that ∠BAD = ∠EAC Adding ∠DAC to both sides, we get ∠BAD + ∠DAC = ∠EAC + ∠DAC => ∠BAC = ∠DAE …(i) Now in ∆BAC and ∆EAD, we have ∠BAC = ∠DAE [using (i)] AB = AD (given) AC = AE (given) ∆BAC ≅ ∆EAD (by SAS congruence rule) => BC = DE (by c.p.c.t)

Answer 2

Answer:

→ Hey Mate,

→ Given Question: - In Fig AC = AE, AB = AD and ∠ BAD = ∠ EAC. prove that BC = DE.

→ To find : Prove that BC  = BC

Step-by-step explanation:

→ solution : Join DE

→ We Have ,

→ ∠BAD = ∠EAC

→ ∠BAD + ∠DAC = ∠EAC + ∠DAC          [ Adding ∠ DAC to both sides ] =(1)

→ Now , in triangles ABC and ADE , we have

→ AB = AD     [ Given]

→ ∠BAC = ∠DAE   [ From ( 1 ) ]

→ AC = AE     [ Given ]

→ so , by SAS congruency criterion , we have

→ Δ ABC ≅ Δ ADE

→ BC = DE   [ cpct ]

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