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

By AA similarity angle GBA=ECB(corresponding angle)

BGA=CGB(right angle)

SO BY AA SIMILARITY

∆ABG~∆DCB

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.

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@GauravSaxena01