Answer:
Answer of 15(i) is the picture . 15(ii) and 17 are solved
Step-by-step explanation:∆ABC where B=90°
→D is mod point of BC so CD=BD
To Prove:
→BC^2=4(AD^2-AB^2)
PROOF:--
In ∆ABD by pythagorus theorem we get,
AD^2=AB^2+BD^2
SO
➡️BD^2= AD^2-AB^2. (1)
NOW ,
BC= BD+CD
BC= 2BD (AS D IS MID POINT)
SQUAREING BOTH SIDES
BC^2= 4BD^2
BY EQUATION 1 WE GOT BD^2 = AD^2-AB^2
Putting this value
➡️BC^2= 4(AD^2-AB^2)
HENCE PROVED
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17.
Construction : Draw AE ⊥ BC
Proof : In ∆ABE and ∆ACE, we have
AB = AC [given]
AE = AE [common]
and ∠AEB = ∠AEC [90°]
Therefore, by using RH congruent condition
∆ABE ~ ∆ACE
⇒ BE = CE
In right triangle ABE.
AB2 = AE2 + BE2 ...(i)
[Using Pythagoras theorem]
In right triangle ADE,
AD2 = AE2 + DE2
[Using Pythagoras theorem]
Subtracting (ii) from (i), we get
AB2 - AD2 = (AE2 + BE2) - (AE2 + DE2)
AB2 - AD2 = AE2 + BE2 - AE2 - DE2
⇒ AB2 - AD2 = BE2 - DE2
⇒ AB2 - AD2 (BE + DE) (BE - DE)
But BE = CE [Proved above]
⇒ AB2 - AD2 = (CE + DE) (BE - DE)
= CD.BD
⇒ AB2 - AD2 = BD.CD
Hence Proved.
