In the given figure, Δ FEC ≅ Δ GDB and ∠ 1 = ∠ 2.Prove that

Q) In the given figure, Δ FEC ≅ Δ GDB and ∠ 1 = ∠ 2. Prove that Δ ADE ~ Δ ABC.

Ans:

Step 1: Since Δ FEC ≅ Δ GDB

∴ BD = CE (by CPCT)

Step 2: Since ∠ 1 = ∠ 2

∴ AD = AE (since sides opposite to equal angles are always equal)

Step 3: In Δ ABC, AB = AD + BD

AB = AE + CE
(AD = AE calculated above in Step 2 and BD = CE calculated above in Step 1)

∵ AE + CE = AC

∴ AB = AC

Step 4: Let’s now take ratio of left sides in Δ ADE and Δ ABC

i.e. $\frac{AD}{AB}$

Since AD = AE calculated above in Step 2

and AB = AC calculated above in Step 3

∴ $\frac{AD}{AB}$ = $\frac{AE}{AC}$

Step 5: Now since $\frac{AD}{AB} = \frac{AE}{AC}$

∴ By Triangles’ symmetry identity, Δ ADE ~ Δ ABC … Hence Proved !

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