Question

15th and 17th one pls 1st Correct answer will be marked as brainiest

Answer 1

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.