हिंदी

In triangle ABC, D and E are points on side AB such that AD = DE = EB. Through D and E, lines are drawn parallel to BC which meet side AC at points F and G respectively.

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प्रश्न

In triangle ABC, D and E are points on side AB such that AD = DE = EB. Through D and E, lines are drawn parallel to BC which meet side AC at points F and G respectively. Through F and G, lines are drawn parallel to AB which meets side BC at points M and N respectively. Prove that: BM = MN = NC.

योग
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उत्तर

The figure is shown below

AD = DE = EB

In ΔAEG

AD = DE & DF || EG

Using mid point theorem

F is midpoint of AG

∴ AF = FG         ...(1)

DF || EG || BC and DE = BE,

∴ FG = GC          ...(2)

(1), (2) we get

AF = GF = GC

Similarly since GN || FM || AB

∴ BM = MN = NC

Hence, proved.

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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 १५४]

APPEARS IN

सेलिना Concise Mathematics [English] Class 9 ICSE
अध्याय 11 Mid-point Theorem and Its Converse [Including Intercept Theorem]
Exercise 12 (B) | Q 7 | पृष्ठ १५४

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