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Basic Proportionality Theorem

In the figure...

Question

In the figure, given below, the medians BD and CE of a triangle ABC meet at G. Prove that : (i) $Δ$ EGD~ $Δ$ CGB and (ii) BG=2 GD from (i) above.

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Solution

(i) Since, BD and CE are medians.

AD = DC

AE = BE

Hence, by converse of Basic Proportionality theorem,

ED || BC

In EGD and CGB,

(ii) Since,

$fraction numerator G D over denominator G B end fraction equals fraction numerator E D over denominator B C end fraction minus negative negative open parentheses 1 close parentheses$

In AED and ABC,

$angle A E D equals angle A B C space space space space open parentheses c o r r e s p o n d i n g space a n g l e s close parentheses angle E A D equals angle B A C space space space space open parentheses c o m m o n close parentheses increment E A D tilde increment B A C space space space space space open parentheses A A space s i m i l a r i t y close parentheses therefore fraction numerator E D over denominator B C end fraction equals fraction numerator A E over denominator A B end fraction equals 1 half space space open parentheses sin c e comma E space i s space t h e space m i d minus p o i n t space o f space A b close parentheses fraction numerator E D over denominator B C end fraction equals 1 half$

From (1),

$fraction numerator G D over denominator G B end fraction equals 1 half G B equals 2 G D$

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Q. In the adjoining figure, medians

A D

C E

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Q. In the adjoining figure, the medians BD and CE of a

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B G = 2 \times G D

Q. In given figure, CD and GH are respectively the medians of

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△ A B C \sim △ F E G

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\frac{C D}{G H} = \frac{A B}{F E}

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In the figure, the medians BD & CE of a ∆ ABC meet at G. Find the ratio of GD : GB.

Q. In given figure, If

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