Question

In given figure, EB ⊥ AC, BG ⊥ AE and CF ⊥ AE (2015)

Prove that:

(a) ∆ABG ~ ∆DCB

(b) bc/bd = be/ba
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Answer 1

hope it helps you......

Answer 2

Answer:

Step-by-step explanation:

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.