Question

In the given figure EB is perpendicular to AC, BG is perpendicular to AE and CF is perpendicular to AE. prove that triangle ABC is similar to triangle DCB also prove that BC/BD = BE/BA

Answer 1

pls post the diagram of this question or recorrect ur question

Answer 2

Solution :-

1/(BD)2= 1/ (BC)2+1/(AB)2

1/ (BD)2=1/ (BC)2+1(AB)2

1/ (BD)2=(AB)2+ (BC)2/ (BC)2× (AB)2

1/ (BD)2=(AB)2+(BC)2/ (BC)2× (AB)2;

Since (AB)^2 + (BC)^2 = (AC)^2, then

1/ (BD)2=(AC)2/ (BC)2× (AB)2

1/ (BD)2 =(AC)2(BC)2× (AB)2;

(BC)2×(AB)2=(AC)2×(BD)2(BC)2×(AB)2

=(AC)2× (BD)2;

(BC×AB)2=(AC×BD)2(BC×AB)2

=(AC×BD)2.

The area of the triangle is 1/2×BC×AB as well as 1/2×AC×BD. Thus BC×AB = AC×BD.

Therefore

(BC×AB)2=(AC×BD)2(BC×AB)2=(AC×BD)2 is true.

==============

@GauravSaxena01