Question

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

Prove that:

(a) ∆ABG ~ ∆DCB

(b) bc/bd = be/ba
​

Answer 1

Answer:

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.

Answer 2

Answer:

${\huge{\underline{\underline{\bold{\red{Answer:-}}}}}}$

☁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.

hope help u mate ✌