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Answer 1

Question:-

12. O is the orthocenter of the triangle ABC Forces P,Q,R acting along OA, OB, OC are in equilibrium.

$\sf \large Prove \: that: \frac{P}{BC} = \frac{Q}{CA} = \frac{R}{AB} .$

Given:-

• O is the orthocenter of the triangle ABC Forces P,Q,R acting along OA, OB, OC are in equilibrium.

To Prove:-

$\sf \large \frac{P}{BC} = \frac{Q}{CA} = \frac{R}{AB} .$

Solution:-

• Let, AD, BE and CF are ⊥ drawn A,B,C.
• Since, O is the Orthocenter, they will meet at O.

Let, ∠BOC = α, ∠COA = β and ∠AOB = θ.

Now, consider ∆ADB and ∆BFC.

∵ ∠ADB = ∠BFC = 90° ; ∠ABD and ∠FBC are common (LB).

Hence, remaining angles ∠BAD = ∠FCB.

Parallely In ∆ADC + ∆BEC, ∠CAD = ∠EBC.

∴ α = ∠BHC = 180° – [ ∠OBC + ∠OCB ]

= 180° – [ ∠EBC + ∠FCB ]

= 180° – [ ∠CAD + ∠BAD ]

= 180° – ∠BAC (or) α = 180° – A.

Parallely β = 180° – B and θ = 180° – C.

as per Lami's Theorem, Forces P,Q,R acting

OA,OB and OC respectively will be equilibrium.

If $\sf \large \frac{P}{sin(180 \degree - \alpha)} = \frac{Q}{sin \beta} = \frac{R}{ sin \theta} \: (or) \: \frac{P}{sin (180 \degree - A} = \frac{Q}{sin(180 \degree -B) } = \frac{R}{sin(180 \degree -C) }$

$\sf \large \therefore \frac{P}{sin \: A} = \frac{Q}{sin \: B} = \frac{R}{si n \: C} \: \: ⇒(1)$

As per Sine Law ⇒ Sin A = ak ; Sin B = bk ; Sin C = ck.

$\sf \large Replace \: in \: equation \: (1)⇒ \frac{P}{ak} = \frac{Q}{bk} = \frac{R}{ck}$

Answer:-

$\sf \large \color{red}Hence, Proved \: that \: \sf \large \color{red} \sf \large \frac{P}{BC} = \frac{Q}{CA} = \frac{R}{AB} .$

Hope you have satisfied. ⚘