Question

In triangke ABC , if D is midpoint of AB and DE is parrlel to BC , then find the ratio of AE : ED.​

Answer 1

Given:

In ΔABC, D is a midpoint of AB

DE // BC

To find:

The ratio of AE : ED

Solution:

We know that,

Converse of Basic Proportionality theorem:- If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

∴ $\frac{AD}{DB} = \frac{AE}{EC}$....... (i)

Since D is a midpoint of AB (given)

∴ AD = DB ...... (ii)

From (i) & (ii), we get

AE = EC

⇒ E is a midpoint of AC ..... (iii)

Now,

Consider Δ AED and ΔACB, we have

∠A = ∠A ........ [common angle]

∠ADE = ∠ABC ..... [corresponding angles, ∵DE//BC and side AB forms a transversal]

∴ Δ AED ~ ΔACB ........ By AA Similarity

Also, we know that the corresponding sides of two similar triangles are proportional to each other.

∴  $\frac{AE}{AC} = \frac{ED}{BC}$

on rearranging, we get

⇒ $\frac{AE}{ED} = \frac{AC}{BC}$

⇒ $\frac{AE}{ED} = \frac{AE \:+\: EC}{BC}$

⇒  AE : ED = (AE+EC) : BC

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Also View:

In the figure,DE ||BC If DB=6cm,AE=2cm,EC=4cm Find the length of AD

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In figure.6.17. (i) and (ii), DE || BC. Find EC in (i) and AD in (ii). (i) (ii)

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In the given figure, DE║BC and DE: BC = 3:5. Calculate the ratio of the areas of ∆ADE and the trapezium BCED.

brainly.in/question/8027386

Answer 2

Given :-

In ΔABC, D is a midpoint of AB

DE // BC

To find :-

The ratio of AE : ED

Solution :-

We know that,

Converse of Basic Proportionality theorem:- If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

∴ AD / DB = AE / EC ....... (i)

Since D is a midpoint of AB (given)

∴ AD = DB ...... (ii)

From (i) & (ii), we get

AE = EC

⇒ E is a midpoint of AC ..... (iii)

Now,

Consider Δ AED and ΔACB, we have

∠A = ∠A ........ [common angle]

∠ADE = ∠ABC ..... [corresponding angles, ∵DE//BC and side AB forms a transversal]

∴ Δ AED ~ ΔACB ........ By AA Similarity

Also, we know that the corresponding sides of two similar triangles are proportional to each other.

∴ AE / AC = ED / BC

on rearranging, we get

⇒ AE / ED = AC / BC

⇒ AE / ED = AE + EC / BC

⇒ AE : ED = (AE+EC) : BC