Question
Download Solution PDFABCD वर्गामध्ये, कर्ण AC आणि BD O येथे छेदतात. ∠CAB चे कोन दुभाजक BD आणि BC ला अनुक्रमे F आणि G येथे भेटतात. OF ∶ CG च्या समान किती आहे?
Answer (Detailed Solution Below)
Detailed Solution
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ABCD एक चौरस आहे
कर्ण O येथे छेदतो
∠CAB चा कोन दुभाजक BD आणि BC ला F आणि G वर भेटतो
गणना:
बाजू AB x असू द्या
तर, कर्ण AC = √2.x
⇒ OA = OB = OC = OD = AC/2 = x/√2 ----(i)
आता, ΔAOB मध्ये आणि कोन दुभाजक प्रमेयातून
AB : AO = BF : OF
⇒ x : (x/√2) = BF/OF
⇒ √2 : 1 = BF/OF
BF = √2y, OF = y ----(ii) समजा
पासून, BF + OF = OB = x/√2 [पासून (i)]
⇒ √2y + y = x/√2
⇒ \(y = \frac{x}{\sqrt2 . (\sqrt2 + 1)}\)
⇒ \(OF = \frac{x}{\sqrt2 . (\sqrt2 + 1)}\) ----(iii) [पासून (ii)]
आता, ΔABC मध्ये आणि कोन दुभाजक प्रमेयातून
AB : AC = BG : GC
⇒ x : √2 x = BG : GC
⇒ 1 : √2 = BG : GC
चला BG = z, GC = √2z ----(iv)
पासून, BG + GC = BC = x
⇒ z + √2z = x
⇒ \(z = \frac{x}{(\sqrt2 + 1)}\)
⇒ \(\sqrt2.z = \frac{\sqrt2x}{(\sqrt2 + 1)}\)
⇒ \(CG = \frac{\sqrt2x}{(\sqrt2 + 1)}\) ----(v )
आता (iii) आणि (v) वरून, आपल्याला मिळेल
OF : CG = \( \frac{x}{\sqrt2 . (\sqrt2 + 1)} : \frac{\sqrt2x}{(\sqrt2 + 1)}\)
⇒ OF : CG = \( \frac{1}{\sqrt2} : \frac{\sqrt2}{1}\)
⇒ OF : CG = 1 : 2
∴ आवश्यक गुणोत्तर 1 : 2 आहे.
Last updated on Jul 19, 2025
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