Question

question 6 plzz....​

Answer 1

Answer:

since triangle ABE ~  triangleACD

Therefore AB = AC (1)

AD = AE (2)

Now, ,in ADE and ABC

Dividing equation (2) by (1)

AD/AB=AE/AC

angle A=angle A(common angle)

therefore ,

triangle ADE  ~  ABC triangle

Step-by-step explanation:

Answer 2

Q6. In fig. 6.37, If ∆ABE ≅ ∆ACD, show that ∆ ADE ≅ ∆ ABC.

Ans:- Since Δ ABE ≅ Δ ACD

Hence, BE = CD

And ∠DBE = ∠ECD

In Δ DBE and Δ ECD

BE = CD (proved earlier)

∠DBE = ∠ECD (proved earlier)

DE = DE (common)

Hence, Δ DBE ≅ Δ ECD

This means; DB = EC

This also means;

AD/DB = AE/EC

Hence; DE || BC

Thus, Δ ADE ∼ Δ ABC proved.

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