Question

In the given figure, AB ⊥ BC, FG ⊥ BC and DE ⊥ AC. Prove that ΔADE ∼ ΔGCF.

Answer 1

Solution:-
In Δ ABC and Δ GCF
∠ ABC = ∠ GFC = 90°
Angle C = Angle C ....(common) 
Therefore, Δ ABC is similar to Δ GCF ....... (1) ..... By AA similarity
Now, 
In Δ ADE and Δ ABC
Angle DEA = Angle ABC = 90°
∠ A = ∠ A (common)
Therefore, Δ ADE is similar to Δ ABC ...... (2) .... By AA similarity
From (1) and (2 )
Δ ADE is similar to Δ GCF
Hence proved.

Answer 2

ANSWER

In the given figure we have to prove Δ ADE and Δ FGC are similar.

Now from Δ ADE and Δ GFC we have,

∠AED=∠GFC [right-angles],

∠DAE=∠FGC [ AD∥GF andAC intersector, similar angles].

Then ∠ADE=∠GCF [Remaining angles].

∴ΔADE is similar to ΔGFC.[By AA ] [henceproved]