Let RB = x

BQR is ext of ΔPBQ

∴ PBQ = 2

Now inΔ PBQ, PBQ = QPB

PQ = QB = d

Also, BRA is ext of Δ BQR

∴ QBR = 3

And BRQ = 3

(Linear Pair)

Now in ΔBQR, by applying Sine Law, we get

d/sin (-3

) = 3d/4 /sin

= x/sin²

d/sin 3

= 3d/4 /sin

= x/sin²

d/3 sin

4 sin³

= 3d/4 sin

= x/2sin

cos

d/3 4sin²

= 3d/4 = x/2cos

..................(I)(II)(III)

From eq. (I),I=II

d/3 4sin²

= 3d/4 ⇒4 = 9 12 sin²

sin²

= 5/12 ⇒cos²

= 7/12

Also from eq. (1) using (II) and (III) we have

3d/4 = x/2 cos

⇒4x²² = 9 d²cos²

x² = 9d²/4 = 7/12 = 21/16 d²...............(3)

Again from ΔABR, we have sin 3

= h/x

3 sin

4 sin³

= h/x ⇒sin

(3 4sin²

) = h/x

sin

(3 4 x5/9) = h/x (using sin²

= 5/12)

4/3 sin

= h/x

Squaring both sides, we get

16/9 sin²

= h²/x²16/9x2/12 = h²/x²

(again using sin²

= 5/12)

h² = 4 x 5/9 x 3 x²⇒ h² = 20/27 x 21/16 d²

{using value of x² from eq.(3)}⇒h² = 35/36 d²⇒36 h² = 35d²

Proved

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