Question

solve the question seen in attachment ​

Answer 1

Answer:

1st question is by SSS congruence rule

Step-by-step explanation:

2nd by cpct

Answer 2

Solution :-

(i) in ∆ABD and ∆CBD

• AD = DC (given)
• AB = BC (given)
• DB = DB (common)

so, ∆ABD ≅ ∆CBD by SSS congruency rule

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now,

∆ABD ≅ ∆CBD then,

⟹ ∠ADB = ∠CDB (By CPCT)

but, ∠ADB + ∠CDB = ∠B

⟹ 2(∠ADB) = ∠B

⟹ ∠ADB = ∠B/2

so, we can say that, BD bisects ∠B

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∆ABD ≅ ∆CBD then,

⟹ ∠ABD = ∠CBD (by CPCT)

but ,∠ABD + ∠CBD = ∠D

⟹ 2(∠ABD) = ∠D

⟹ ∠ABD = ∠D/2

so, we can say that, BD bisects ∠D

hence, proved

that, BD bisects ∠B as well as ∠D