Question

In the figure,AB perpendicular to BC. DE perpendicular to AC & GF perpendicular to BC. Show that ∆ADE ~ ∆GCF

Answer 1

Answer:

∆ADE ~ ∆GCF is proved as follows:

In △ ABC and △ ADE,

∠ A = ∠ A Similar.

∠ DAE = ∠ BAC

△ ABC and △ AED are similar.

In triangle ABC and triangle CFD,

∠C = ∠C Similar.

∠ GFC = ∠ABC

By AA criterion for similarity,

△ ABC and  △ GFC are similar.

Since △  ADE and △ GCF is similar to △ ABC,

△  ADE and  △ GCF are similar.

Hence, Proved.

Answer 2

Answer:

∆ADE ~ ∆GCF is proved as follows

Step-by-step explanation:

In △ ABC and △ ADE,

∠ A = ∠ A Similar.

∠ DAE = ∠ BAC

△ ABC and △ AED are similar.

In triangle ABC and triangle CFD,

∠C = ∠C Similar.

∠ GFC = ∠ABC

By AA criterion for similarity,

△ ABC and  △ GFC are similar.

Since △  ADE and △ GCF is similar to △ ABC,

△  ADE and  △ GCF are similar.

Hence, Proved.