Question

Cd and gh are respectively the bisectors of ∠ acb and ∠ egf such that d and h lie on sides ab and fe of δ abc and δ efg respectively. If δ abc ~ δ feg, show that: (i) cd ac  gh fg (ii) δ dcb ~ δ hge (iii) δ dca ~ δ hgf

Answer 1

Sin^2 A = 1/2 tan^2A= 1/(2*(1)^2) =1/2, As angle A is an acute angle.

So,

SinA = 1/√2 , You know value of sin 45°=1/√2

Or,

A = Sin^(-1) {1/√2} = 45°

So, A = 45°

Answer 2

Given, CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ΔABC and ΔEFG respectively.  

(i) From the given condition,

ΔABC ~ ΔFEG.

∴ ∠A = ∠F, ∠B = ∠E, and ∠ACB = ∠FGE  

Since, ∠ACB = ∠FGE

∴ ∠ACD = ∠FGH (Angle bisector)

And, ∠DCB = ∠HGE (Angle bisector)  

In ΔACD and ΔFGH,

∠A = ∠F  

∠ACD = ∠FGH  

∴ ΔACD ~ ΔFGH (AA similarity criterion)

⇒CD/GH = AC/FG  

(ii) In ΔDCB and ΔHGE,  

∠DCB = ∠HGE (Already proved)

∠B = ∠E (Already proved)  

∴ ΔDCB ~ ΔHGE (AA similarity criterion)

(iii) In ΔDCA and ΔHGF,

∠ACD = ∠FGH (Already proved)

∠A = ∠F (Already proved)

∴ ΔDCA ~ ΔHGF (AA similarity criterion)