Question

In the fig., AD = 3, DB = 12, AE = 4 and EC = 16.
Prove that BC = 5DE​

Answer 1

Solution :-

→ AD/DB = 3/12 = 1/4

So,

→ AD/(AD + DB) = 1/(1 + 4)

→ AD/AB = 1/5 ----- Eqn.(1)

similarly,

→ AE/EC = 4/16 = 1/4

So,

→ AE/(AE + EC) = 1/(1 + 4)

→ AE/AC = 1/5 ------- Eqn.(2)

then, from Eqn.(1) and Eqn.(2)

→ AD/AB = AE/AC

Now, according to the converse of BPT :-

• If a line segment is drawn to cut two sides of a triangle in same proportion, then it is parallel to the third side .

therefore, we can conclude that,

→ DE || BC .

now, in ∆ADE and ∆ABC we have,

→ ∠DAE = ∠BAC { common . }

→ ∠ADE = ∠ABC { Corresponding angles. }

So,

→ ∆ADE ~ ∆ABC { By AA similarity. }

then,

→ AD/AB = AE/AC = DE/BC { when two ∆'s are similar their corresponding sides are in same proportion. }

hence, from Eqn.(1) we get,

→ 1/5 = DE/BC

→ BC = 5•DE (Proved)

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