Question

In the given figure EB|AC,BG|AE and CF|AE prove that BC/BD=BE/BA

Answer 1

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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