Question

In ∆ ABC and ∆EDC, AB is parallel to ED. BD = 1
/3BC and AB = 12.3 cm.

(i) Prove that ∆ABC ~∆EDC.

(ii) Find DE
.

Answer 1

In the attachment I have solved this problem.

I applied basic proportionality theorem if AB is parallel to ED

I hope this answer help you

Answer 2

Answer:

The ans is 8.2 unit

Step-by-step explanation:

as per the question we prove that Prove that ∆ABC ~∆EDC.

in  ∆ABC and ∆EDC.

Given is AB ║ED

Then we know that $\frac{CE}{EA} =\frac{CD}{DB}$

Then  ∠c=∠c

∠CAB = ∠CDE

By corrspoding angel

now by AAA theorem we say that

∆ABC ~∆EDC.

Now we find the DE

$\frac{CE}{CA} =\frac{CD}{CB} =\frac{ED}{AB}$= c0nstant

$\frac{CD}{BD} =\frac{2x}{X}=2$

now ED= 2*12.3/3=8.2 unit

Hence the ans is 8.2 unit.

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