Triangle DEF is right-angled at E, EG is perpendicular to DF and GH i

1.	Triangle DEF is right-angled at E, EG is perpendicular to DF and GH is perpendicular to EF. Prove that:1) GH2=EH.HF2) EG2=DG.GF
Answer» Given :- ∠E = 90° . EG ⟂ DF . GH ⟂ EF. To Prove :- (i) GH² = EH * HF (ii) EG² = DG * GF . Solution :- Let us assume that, ∠F = x° . Than, → ∠FEG = (90 - x°) { As, EG ⟂ DF .} Also, in ∆GHF, → ∠HGF = (90 - x°) { As, GH ⟂ EF. } Now, in ∆GHE and ∆FHG, we have, → ∠GHE = ∠FHG (90°) → ∠GEH = ∠FGH (90 - x°) So, → ∆GHE ~ ∆FHG . (By AA similarity) . Therefore, → GH/EH = HF/GH (By CPCT) → *GH² = EH HF (Proved). ________________________________________________________________________ Now, Similarly, → ∠EFD = x° → ∠EDG = (90 - x°) → ∠FEG = (90 - x°) Now, in ∆EGF and ∆DGE, → ∠EGF = ∠DGE. (90°) → ∠FEG = ∠EDG. (90 - x°) So, → ∆EGF ~ ∆DGE .(By AA Similarity.) Therefore, → EG / GF = DG / EG (By CPCT. ) Hence*, → EG² = DG GF. (Proved.)

Triangle DEF is right-angled at E, EG is perpendicular to DF and GH is perpendicular to EF. Prove that:1) GH2=EH.HF2) EG2=DG.GF

Answer»

Given :-

To Prove :-

Solution :-

Let us assume that, ∠F = x° .

Than,

→ ∠FEG = (90 - x°) { As, EG ⟂ DF .}

Also, in ∆GHF,

→ ∠HGF = (90 - x°) { As, GH ⟂ EF. }

Now, in ∆GHE and ∆FHG, we have,

→ ∠GHE = ∠FHG (90°)

→ ∠GEH = ∠FGH (90 - x°)

So,

→ ∆GHE ~ ∆FHG . (By AA similarity) .

Therefore,

→ GH/EH = HF/GH (By CPCT)

→ GH² = EH * HF (Proved).

________________________________________________________________________

Now, Similarly,

→ ∠EFD = x°

→ ∠EDG = (90 - x°)

→ ∠FEG = (90 - x°)

Now, in ∆EGF and ∆DGE,

→ ∠EGF = ∠DGE. (90°)

→ ∠FEG = ∠EDG. (90 - x°)

So,

→ ∆EGF ~ ∆DGE .(By AA Similarity.)

Therefore,

→ EG / GF = DG / EG (By CPCT. )

Hence,

→ EG² = DG * GF. (Proved.)