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In a triangle ABC, AD is a median and E is mid-point of median AD. A line through B and E meets AC at point F. Prove that: AC = 3AF.

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प्रश्न

In a triangle ABC, AD is a median and E is mid-point of median AD. A line through B and E meets AC at point F.

Prove that: AC = 3AF.

योग
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उत्तर

The required figure is shown below

For help, we draw a line DG || BF

Now from triangle ADG, DG || BF and E is the midpoint of AD

Therefore, F is the midpoint of AG, i.e; AF = FG        ...(1)

From triangle BCF, DG || BF and D is the midpoint of BC

Therefore, G is the midpoint of CF, i.e; FG = GC       …(2)

AC = AF + FG + GC

= AF + AF + AF

AC = 3AF                   ...(From (1) and (2))

Hence, proved.

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 12: Mid-point and Its Converse [ Including Intercept Theorem] - Exercise 12 (A) [पृष्ठ १५१]

APPEARS IN

सेलिना Concise Mathematics [English] Class 9 ICSE
अध्याय 12 Mid-point and Its Converse [ Including Intercept Theorem]
Exercise 12 (A) | Q 13 | पृष्ठ १५१

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