Question

If in two triangles, corresponding angles are equal, then their
corresponding sides are in the same ratio (or proportion) and hence the two
triangles are similar.

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Answer 1

Step-by-step explanation:

Given: Two triangles ∆ABC and ∆DEF such that ∠A = ∠D, ∠B = ∠E & ∠C = ∠F

To Prove: ∆ABC ~ ∆DEF

Construction: Draw P and Q on DE & DF such that DP = AB and DQ = AC respectively and join PQ.

Proof: In ∆ABC and ∆DPQ

AB = DP

∠A = ∠D

AC= DQ

⇒ ∆ABC ≅ ∆DPQ

⇒ ∠B = ∠P

But, ∠B = ∠E [Given]

Thus, ∠P = ∠E

For lines PQ & EF with transversal PE, ∠ P & ∠ E are corresponding angles, and they are equal Hence, PQ ∥ EF.

Now,

Since, PQ ∥ EF.

We know that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

Hence, DP/PE = DQ/QF

PE/DP = QF/DQ

Adding 1 on both sides,

PE/DP + 1 = QF/DQ+ 1

(PE + DP)/DP = (QF + DQ)/DQ

=>DE/DP = DF/DQ

⇒ DP/DE = DQ/DF

And by construction,

DP = AB and DQ = AC

⇒ AB/DE = AC/DF

Similarly, we can prove that AB/DF = BC/EF

Therefore, AB/DE = AC/DF = BC/EF

Since all 3 sides are in proportion

∴ ∆ABC ~ ∆DEF

Hence Proved.