Question

1. In △ ABC, AB=AC and D is any point on BC. DE and DF are perpendiculars drawn on AB and AC respectively and BG is perpendicular to AC. Prove that DE + DF = BG....Please hurry...​

Answer 1

ANSWER

In triangle ABC

D is the midpoint of BC

DE perpendicular to AB

And DF perpendicular to AC

DE=DF

To prove:

Triangle ABC is an isosceles triangle

Proof:

In the right angles triangle BED and CDF

Hypotenuse BD=DC ( because D is a midpoint )

Side DF=DE ( given)

△BED≅CDF ( RHS axiom)

∠C=∠B

AB=AC ( sides opposite to equal angles

△ABC is an isosceles triangle

Answer 2

Answer:

triangle ABC

D is the midpoint of BC

DE perpendicular to AB

And DF perpendicular to AC

DE=DF

To prove:

Triangle ABC is an isosceles triangle

Proof:

In the right angles triangle BED and CDF

Hypotenuse BD=DC ( because D is a midpoint )

Side DF=DE ( given)

△BED≅CDF ( RHS axiom)

∠C=∠B

AB=AC ( sides opposite to equal angles

△ABC is an isosceles triangle