हिंदी

The side AC of a triangle ABC is produced to point E so that CE = AC. D is the mid-point of BC and ED produced meets AB at F.

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प्रश्न

The side AC of a triangle ABC is produced to point E so that CE = AC. D is the mid-point of BC and ED produced meets AB at F. Lines through D and C are drawn parallel to AB which meet AC at point P and EF at point R respectively.

Prove that:

  1. 3DF = EF
  2. 4CR = AB
योग
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उत्तर

Consider the figure :

Here D is the midpoint of BC and DP is parallel to AB, therefore P is the midpoint of AC and 

PD = `[1]/[2]`AB

(i) Again from the triangle AEF, we have AF || PD || CR and AP = `[1]/[3]`AE

Therefore, DF = `[1]/[3]` EF or we can say that 3DF = EF.

Hence, it is shown.

(ii) From the triangle PED, we have PD || CR and C is the midpoint of PE, therefore, CR = `[1]/[2]`PD

Now,

PD = `1/2` AB

`1/2"PD" = 1/4`AB

CR = `1/4`AB

4CR = AB

Hence, it is shown.

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 11: Mid-point Theorem and Its Converse [Including Intercept Theorem] - Exercise 12 (B) [पृष्ठ १५३]

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सेलिना Concise Mathematics [English] Class 9 ICSE
अध्याय 11 Mid-point Theorem and Its Converse [Including Intercept Theorem]
Exercise 12 (B) | Q 3 | पृष्ठ १५३
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