त्रिभुजों की समरूपता - Revision history

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	== प्रमेय 2 ==		== प्रमेय 2 ==
	~~If a line divides any two sides of a triangle in the same ratio~~, ~~then the line is parallel to the third side.~~		यदि कोई रेखा त्रिभुज की किन्हीं दो भुजाओं को समान अनुपात में विभाजित करती है, तो वह रेखा तीसरी भुजा के समांतर होती है।
	[[File:Triangle - Theorem 2.jpg\|alt=Fig.2\|thumb\|चित्र .2 ]]		[[File:Triangle - Theorem 2.jpg\|alt=Fig.2\|thumb\|चित्र .2 ]]



	~~Let~~ <math>ABC</math> ~~be a triangle with sides~~ <math>AB,AC, BC</math>. <math>DE</math> ~~divides any two sides of the triangle in the same ratio.~~		मान लीजिए <math>ABC</math> एक त्रिभुज है जिसकी भुजाएँ <math>AB,AC, BC</math> हैं। <math>DE</math> त्रिभुज की किन्हीं दो भुजाओं को समान अनुपात में विभाजित करता है।

	'''~~Given~~''': ~~Line divides a triangle in the same ratio. Thus~~ <math>\frac{AD}{DB}=\frac{AE}{EC}</math>		'''दिया हुआ''': रेखा त्रिभुज को समान अनुपात में विभाजित करती है। इस प्रकार <math>\frac{AD}{DB}=\frac{AE}{EC}</math>

	'''प्रमाण:'''		'''प्रमाण:'''

	<math>\frac{AD}{DB}=\frac{AE}{EC}....(1)</math> (~~Given~~)		<math>\frac{AD}{DB}=\frac{AE}{EC}....(1)</math> (दिया हुआ)

	~~Let us assume that~~ <math>DE</math> ~~is not parallel to~~ <math>BC</math>~~. Now we draw~~ <math>DE^' \parallel BC</math>		आइये मान लेते हैं <math>DE</math>, <math>BC</math> के समांतर नहीं है। अब हम <math>DE^' \parallel BC</math> बनाते हैं

	So		इसलिए

	<math>\frac{AD}{DB}=\frac{AE^'}{E^'C}....(2)</math> (~~Property of similar triangles~~)		<math>\frac{AD}{DB}=\frac{AE^'}{E^'C}....(2)</math> (समरूप त्रिभुजों का गुणधर्म)

	~~Therefore, from~~ <math>(1)(2)</math>		अत: <math>(1)(2)</math> से

	<math>\frac{AE}{EC}=\frac{AE^'}{E^'C}</math>		<math>\frac{AE}{EC}=\frac{AE^'}{E^'C}</math>

	~~Now we add~~ 1 ~~to both sides~~,		अब हम दोनों पक्षों में 1 जोड़ते हैं,

	<math>\frac{AE}{EC}+1=\frac{AE^'}{E^'C}+1</math>		<math>\frac{AE}{EC}+1=\frac{AE^'}{E^'C}+1</math>
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	<math>\frac{AE+EC}{EC}=\frac{AE^'+E^'C}{E^'C}</math>		<math>\frac{AE+EC}{EC}=\frac{AE^'+E^'C}{E^'C}</math>

	~~According to the figure~~		चित्र के अनुसार

	<math>AE+EC=AC</math>		<math>AE+EC=AC</math>
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	<math>AE^'+E^'C= AC</math>		<math>AE^'+E^'C= AC</math>

	~~substituting these values in the equation above~~:		उपरोक्त समीकरण में इन मानों को प्रतिस्थापित करने पर:

	<math>\frac{AC}{EC}=\frac{AC}{E^'C}</math>		<math>\frac{AC}{EC}=\frac{AC}{E^'C}</math>

	~~This directly implies that~~		इसका सीधा तात्पर्य यह है कि

	<math>EC=E^'C</math>		<math>EC=E^'C</math>

	~~and~~ <math>E=E^'</math> ~~meaning that they are the same point.~~		और <math>E=E^'</math>जिसका अर्थ है कि वे एक ही बिंदु हैं।

	~~Hence~~ <math>DE</math> ~~is parallel to~~ <math>BC</math>~~. This proves the similarity of triangles.~~		अतः <math>DE</math>, <math>BC</math> के समांतर है। इससे त्रिभुजों की समरूपता सिद्ध होती है।

	[[Category:त्रिभुज]][[Category:गणित]][[Category:कक्षा-10]]		[[Category:त्रिभुज]][[Category:गणित]][[Category:कक्षा-10]]

Mani: added content

2024-09-22T03:05:00Z

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← Older revision		Revision as of 08:35, 22 September 2024
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	== प्रमेय 1 ==		== प्रमेय 1 ==
	[[File:Triangle - theorem -1.jpg\|alt=Fig. 1\|thumb\|200x200px\|चित्र .1]]		[[File:Triangle - theorem -1.jpg\|alt=Fig. 1\|thumb\|200x200px\|चित्र .1]]
	~~If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points~~, ~~the other two sides are divided in the same~~		यदि किसी त्रिभुज की एक भुजा के समांतर एक रेखा खींची जाए जो अन्य दो भुजाओं को भिन्न-भिन्न बिंदुओं पर प्रतिच्छेद करे, तो अन्य दो भुजाएं समान अनुपात में विभाजित हो जाती हैं।

	~~ratio.~~		'''प्रमाण:''' हमें एक <math>\triangle ABC</math> दिया गया है जिसमें भुजा <math>BC</math> के समांतर एक रेखा अन्य दो भुजाओं <math>AB</math> और <math>AC</math> को क्रमशः <math>D</math> और <math>E</math> पर प्रतिच्छेद करती है (चित्र-1 देखें)।

	~~'''Proof :''' We are given a~~ <math>\~~triangle ABC~~</math> ~~in which a line parallel to side <math>BC</math> intersects other two sides <math>AB</math> and~~		हमें यह प्रमाणित करने की आवश्यकता है <math>\frac{AD}{DB}=\frac{AE}{EC}</math> .

	<math>AC</math> at <math>D</math> ~~and~~ <math>E</math> ~~respectively (see Fig.1).~~		आइए <math>BE</math> और <math>CD</math> को मिलाएँ और फिर <math>DN \perp AC</math> और <math>EM \perp AB</math> बनाएँ

	~~We need to prove that~~ <math>\~~frac{AD}{DB}~~=\frac{AE}{EC}</math> .		अब, क्षेत्रफल <math>\triangle ADE = \frac{1}{2} AD \times EM</math>

	~~Let us join~~ <math>~~BE</math> and <math>CD</math> and then draw <math>DN~~ \~~perp AC</math> and <math>~~EM ~~\perp AB~~</math>		क्षेत्रफल <math>\triangle BDE = \frac{1}{2} DB \times EM</math>

	~~Now, area of~~ <math>\triangle ADE = \frac{1}{2} AD \times EM</math>		क्षेत्रफल <math>\triangle ADE = \frac{1}{2} AE \times DN</math>

	~~area of~~ <math>\triangle ~~BDE~~ = \frac{1}{2} DB \times EM</math>		क्षेत्रफल <math>\triangle DEC = \frac{1}{2} EC \times DN</math>

	~~area of~~ <math>\~~triangle~~ ADE = \frac{1}{2} AE \times DN</math>		क्षेत्रफल <math>\frac{ADE}{BDE} = \frac{\frac{1}{2}AD \times EM}{\frac{1}{2}DB \times EM}=\frac{AD}{DB} ....(1) </math>

	~~area of~~ <math>\~~triangle~~ DEC = \frac{1}{2} EC \times DN</math>		क्षेत्रफल <math>\frac{ADE}{DEC} = \frac{\frac{1}{2}AE \times DN}{\frac{1}{2}EC \times DN}=\frac{AE}{EC}....(2) </math>

	~~area of~~ <math>\~~frac{ADE}{~~BDE~~} =~~ \~~frac{\frac{1}{2}AD \times EM}{\frac{1}{2}DB \times EM}=\frac{AD}{DB} ....(1)~~ </math>		ध्यान दें कि <math>\triangle BDE</math> और <math>\triangle DEC</math> एक ही आधार <math>DE</math> पर और एक ही समान्तर रेखाओं <math>BC</math> और <math>DE</math> के बीच में हैं,

	~~area of~~ <math>\~~frac{ADE}{DEC}~~ = \~~frac{\frac{1}{2}AE \times DN}{\frac{1}{2}EC \times DN}=\frac{AE}{EC}~~....(2) </math>		क्षेत्रफल <math>\triangle BDE =</math> <math>\triangle DEC....(3)</math>

	~~Note that <math>\triangle BDE</math> and <math>\triangle DEC</math> are on the same base <math>DE</math> and between the same parallels~~		अतः <math>(1) (2) (3)</math> से हमें प्राप्त होता है

	~~<math>BC</math> and <math>DE</math>~~

	~~area of <math>\triangle BDE =</math> area of <math>\triangle DEC....(3)</math>~~

	~~Therefore from~~ <math>(1) (2) (3)</math> ~~we have~~

	<math>\frac{AD}{DB}=\frac{AE}{EC}</math>		<math>\frac{AD}{DB}=\frac{AE}{EC}</math>
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	'''Given''': Line divides a triangle in the same ratio. Thus <math>\frac{AD}{DB}=\frac{AE}{EC}</math>		'''Given''': Line divides a triangle in the same ratio. Thus <math>\frac{AD}{DB}=\frac{AE}{EC}</math>

	'''~~Proof~~:'''		'''प्रमाण:'''

	<math>\frac{AD}{DB}=\frac{AE}{EC}....(1)</math> (Given)		<math>\frac{AD}{DB}=\frac{AE}{EC}....(1)</math> (Given)

Mani: added content

2024-09-22T02:52:01Z

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	~~The similarity of triangles can be defined based on their properties. Two triangles are similar~~, if		त्रिभुजों की समरूपता को उनके गुणों के आधार पर परिभाषित/निर्धारित किया जा सकता है।दो त्रिभुज समरूप होते हैं, यदि

	(i) ~~their corresponding angles are equal and~~		(i) उनके संगत कोण समान हैं तथा

	(ii) ~~their corresponding sides are in the same ratio~~ (~~or proportion~~).		(ii) उनकी संगत भुजाएँ समान अनुपात (या समानुपात) में हैं।

	~~These two are the main conditions for the similarity of triangles.~~		त्रिभुजों की समरूपता के लिए ये दोनों मुख्य स्थितियाँ हैं।

	~~Note that if corresponding angles of two triangles are equal~~, ~~then they are known as equiangular triangles. A famous Greek mathematician Thales gave an important truth relating~~		ध्यान दें कि यदि दो त्रिभुजों के संगत कोण समान हों, तो उन्हें समकोण त्रिभुज कहते हैं। प्रसिद्ध यूनानी गणितज्ञ थेल्स ने दो समकोण त्रिभुजों के संबंध में एक महत्वपूर्ण सत्य बताया जो इस प्रकार है:

	~~to two equiangular triangles which is as follows:~~		दो समकोण त्रिभुजों में किन्हीं दो संगत भुजाओं का अनुपात प्रायः समान होता है।

	~~The ratio of any two corresponding sides in two equiangular triangles is always the same.~~		ऐसा माना जाता है कि उन्होंने इसके लिए मूल आनुपातिकता प्रमेय (जिसे अब थेल्स प्रमेय के रूप में जाना जाता है) नामक परिणाम का उपयोग किया था।

	~~It is believed that he had used a result called the Basic Proportionality Theorem (now known as the Thales Theorem) for the same.~~		== प्रमेय 1 ==
			[[File:Triangle - theorem -1.jpg\|alt=Fig. 1\|thumb\|200x200px\|चित्र .1]]
	== ~~Theorem~~ 1 ==
	[[File:Triangle - theorem -1.jpg\|alt=Fig. 1\|thumb\|200x200px\|~~Fig~~. 1]]
	If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same		If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same

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	<math>\frac{AD}{DB}=\frac{AE}{EC}</math>		<math>\frac{AD}{DB}=\frac{AE}{EC}</math>

	== ~~Theorem~~ 2 ==		== प्रमेय 2 ==
	If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.		If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
	[[File:Triangle - Theorem 2.jpg\|alt=Fig.2\|thumb\|~~Fig~~.2]]		[[File:Triangle - Theorem 2.jpg\|alt=Fig.2\|thumb\|चित्र .2 ]]

Mani at 02:41, 22 September 2024

2024-09-22T02:41:58Z

← Older revision		Revision as of 08:11, 22 September 2024
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			The similarity of triangles can be defined based on their properties. Two triangles are similar, if

			(i) their corresponding angles are equal and

			(ii) their corresponding sides are in the same ratio (or proportion).

			These two are the main conditions for the similarity of triangles.

			Note that if corresponding angles of two triangles are equal, then they are known as equiangular triangles. A famous Greek mathematician Thales gave an important truth relating

			to two equiangular triangles which is as follows:

			The ratio of any two corresponding sides in two equiangular triangles is always the same.

			It is believed that he had used a result called the Basic Proportionality Theorem (now known as the Thales Theorem) for the same.

			== Theorem 1 ==
			[[File:Triangle - theorem -1.jpg\|alt=Fig. 1\|thumb\|200x200px\|Fig. 1]]
			If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same

			ratio.

			'''Proof :''' We are given a <math>\triangle ABC</math> in which a line parallel to side <math>BC</math> intersects other two sides <math>AB</math> and

			<math>AC</math> at <math>D</math> and <math>E</math> respectively (see Fig.1).

			We need to prove that <math>\frac{AD}{DB}=\frac{AE}{EC}</math> .

			Let us join <math>BE</math> and <math>CD</math> and then draw <math>DN \perp AC</math> and <math>EM \perp AB</math>

			Now, area of <math>\triangle ADE = \frac{1}{2} AD \times EM</math>

			area of <math>\triangle BDE = \frac{1}{2} DB \times EM</math>

			area of <math>\triangle ADE = \frac{1}{2} AE \times DN</math>

			area of <math>\triangle DEC = \frac{1}{2} EC \times DN</math>

			area of <math>\frac{ADE}{BDE} = \frac{\frac{1}{2}AD \times EM}{\frac{1}{2}DB \times EM}=\frac{AD}{DB} ....(1) </math>

			area of <math>\frac{ADE}{DEC} = \frac{\frac{1}{2}AE \times DN}{\frac{1}{2}EC \times DN}=\frac{AE}{EC}....(2) </math>

			Note that <math>\triangle BDE</math> and <math>\triangle DEC</math> are on the same base <math>DE</math> and between the same parallels

			<math>BC</math> and <math>DE</math>

			area of <math>\triangle BDE =</math> area of <math>\triangle DEC....(3)</math>

			Therefore from <math>(1) (2) (3)</math> we have

			<math>\frac{AD}{DB}=\frac{AE}{EC}</math>

			== Theorem 2 ==
			If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
			[[File:Triangle - Theorem 2.jpg\|alt=Fig.2\|thumb\|Fig.2]]



			Let <math>ABC</math> be a triangle with sides <math>AB,AC, BC</math>. <math>DE</math> divides any two sides of the triangle in the same ratio.

			'''Given''': Line divides a triangle in the same ratio. Thus <math>\frac{AD}{DB}=\frac{AE}{EC}</math>

			'''Proof:'''

			<math>\frac{AD}{DB}=\frac{AE}{EC}....(1)</math> (Given)

			Let us assume that <math>DE</math> is not parallel to <math>BC</math>. Now we draw <math>DE^' \parallel BC</math>

			So

			<math>\frac{AD}{DB}=\frac{AE^'}{E^'C}....(2)</math> (Property of similar triangles)

			Therefore, from <math>(1)(2)</math>

			<math>\frac{AE}{EC}=\frac{AE^'}{E^'C}</math>

			Now we add 1 to both sides,

			<math>\frac{AE}{EC}+1=\frac{AE^'}{E^'C}+1</math>

			<math>\frac{AE}{EC}+\frac{EC}{EC}=\frac{AE^'}{E^'C}+\frac{E^'C}{E^'C}</math>

			<math>\frac{AE+EC}{EC}=\frac{AE^'+E^'C}{E^'C}</math>

			According to the figure

			<math>AE+EC=AC</math>

			<math>AE^'+E^'C= AC</math>

			substituting these values in the equation above:

			<math>\frac{AC}{EC}=\frac{AC}{E^'C}</math>

			This directly implies that

			<math>EC=E^'C</math>

			and <math>E=E^'</math> meaning that they are the same point.

			Hence <math>DE</math> is parallel to <math>BC</math>. This proves the similarity of triangles.

	[[Category:त्रिभुज]][[Category:गणित]][[Category:कक्षा-10]]		[[Category:त्रिभुज]][[Category:गणित]][[Category:कक्षा-10]]
	~~Similarity of Triangles~~

Mani at 08:18, 23 August 2023

2023-08-23T08:18:25Z

← Older revision		Revision as of 13:48, 23 August 2023
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	[[Category:त्रिभुज]][[Category:गणित]][[Category:कक्षा-10]]		[[Category:त्रिभुज]][[Category:गणित]][[Category:कक्षा-10]]
			Similarity of Triangles

Mani: Updated Category

2023-08-04T12:45:00Z

Updated Category

← Older revision		Revision as of 18:15, 4 August 2023
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	~~[[Category:ज्यामिति]]~~
	[[Category:त्रिभुज]][[Category:गणित]][[Category:कक्षा-10]]		[[Category:त्रिभुज]][[Category:गणित]][[Category:कक्षा-10]]

Sarika at 05:35, 4 August 2023

2023-08-04T05:35:23Z

← Older revision		Revision as of 11:05, 4 August 2023
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	[[Category:ज्यामिति]]		[[Category:ज्यामिति]]
	[[Category:त्रिभुज]][[Category:गणित]]		[[Category:त्रिभुज]][[Category:गणित]][[Category:कक्षा-10]]

Sarika at 05:31, 4 August 2023

2023-08-04T05:31:37Z

← Older revision		Revision as of 11:01, 4 August 2023
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	[[Category:ज्यामिति]]		[[Category:ज्यामिति]]
	[[Category:त्रिभुज]]		[[Category:त्रिभुज]][[Category:गणित]]

Sarika at 10:58, 18 April 2023

2023-04-18T10:58:12Z

← Older revision		Revision as of 16:28, 18 April 2023
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	~~[[Category:गणित]]~~
	[[Category:ज्यामिति]]		[[Category:ज्यामिति]]
	[[Category:त्रिभुज~~]][[Category: कक्षा-10~~]]		[[Category:त्रिभुज]]

Sarika at 06:41, 21 March 2023

2023-03-21T06:41:40Z

← Older revision		Revision as of 12:11, 21 March 2023
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	[[Category:गणित]]		[[Category:गणित]]
	[[Category:ज्यामिति]]		[[Category:ज्यामिति]]
	[[Category:त्रिभुज]]		[[Category:त्रिभुज]][[Category: कक्षा-10]]