B, BC = aSuppose BD = x then DC = a – XAS we know,⇒ AB/AC = BD/DC (angle bisect THEOREM)⇒ c/b = x/(a – x)⇒ ca – CX = bx⇒ x(c + b) = ac⇒ x = ac/(b + c)

"> B, BC = aSuppose BD = x then DC = a – XAS we know,⇒ AB/AC = BD/DC (angle bisect THEOREM)⇒ c/b = x/(a – x)⇒ ca – CX = bx⇒ x(c + b) = ac⇒ x = ac/(b + c)

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In ΔABC, AD bisects ∠A and intersects BC at D. If BC = a, AC = b and AB = c, then BD = ?

SRMJEEE Analytical Geometry in SRMJEEE 11 months ago

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Given, AB = c, AC = B, BC = aSuppose BD = x then DC = a – XAS we know,⇒ AB/AC = BD/DC (angle bisect THEOREM)⇒ c/b = x/(a – x)⇒ ca – CX = bx⇒ x(c + b) = ac⇒ x = ac/(b + c)

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Posted on 08 Nov 2024, this text provides information on SRMJEEE related to Analytical Geometry in SRMJEEE. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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