Question

E is a point on hypotenuse DF of triangle DFH such that segHEperpendicularsegDF, segEG perpendicular seg FH, seg EK perpendicular seg DH. Prove that
1.EG^2=FG ×EK
2.EK^2=DK×EG.
PLEASE GIVE STEP BY STEP EXPLANATION WHOEVER GIVES ANSWER WILL BE MARKED AS BRAINLIEST.​

Answer 1

Step-by-step explanation:

GIVEN

Right triangle DFH with hypotenus DF having points E, G, K on DF, FH, HD such that HE⊥DF, EG⊥FH, EK⊥DH

TO PROVE

1. EG² = FG x EK

2. EK² = DK x EG

ʀᴇʟᴀᴛɪᴏɴ ʙᴇᴛᴡᴇᴇɴ ᴀɴɢʟᴇs

⇒ Since HE⊥DF, ΔEFH is right triangle.

So, ∠F + ∠FHE = 90

Also ∠H = 90

So, ∠FHE + ∠EHK = 90

Substracting the two equation, we get

∠F = ∠EHK ---------------(1)

Similarly

∠D = ∠EHG----------------(2)

⇒ Also

∠FEG + ∠GEH = 90

∠GEH + ∠HEK = 90

Substracting the two equations we get

∠FEG = ∠HEK--------------(3)

Similarly

∠DEK = ∠HEG--------------(4)

PROOF

1. In Δs EFG and EHK

∠F = ∠EHK (by equn (1))

∠FEG = ∠HEK( by equn (3))

∠FGE = ∠HKE( = 90)

Hence By AAA similarity

ΔEFG ~ ΔEHK

So FG/HK = EG/EK

or FG/EG = EG/EK (since EG = HK)

or EG² = FG x EK

2. In Δs EKD and EGH

∠D = ∠EHG(by equn(2))

∠DEK = ∠HEG( by equn 4

∠EKD = ∠EGH( = 90)

Hence By AAA similarity

ΔEKD ~ ΔEGH

So EG/EK = GH/DK

or EG/EK = EK/DK(since EK = GH)

or EK² = DK x EG

Hence Proved.

Answer 2

Answer:

hence proved

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