In the following figure, AD and CE are medians of ∆ABC. DF is drawn parallel to CE. Prove that: (i) EF = FB, (ii) AG: GD = 2: 1 - Noon Academy

In the following figure, AD and CE are medians of ∆ABC. DF is drawn parallel to CE. Prove that: (i) EF = FB, (ii) AG: GD = 2: 1

Class 10 | Exercise 15E | ICSE | Maths | Selina | Similarity (With Applications to Maps and Models)

In the following figure, AD and CE are medians of ∆ABC. DF is drawn parallel to CE. Prove that: (i) EF = FB, (ii) AG: GD = 2: 1

Selina Solutions Concise Class 10 Maths Chapter 15 ex. 15(E) - 14

Solution:

(i) In

$\vartriangle BFD\text{ }and\text{ }\vartriangle BEC,$

$\angle BFD\text{ }=\angle BEC$

[Corresponding angles]

$\angle FBD\text{ }=\angle EBC$

[Common]

Hence,

$\vartriangle BFD\text{ }\sim\text{ }\vartriangle BEC\text{ }by\text{ }AA$

criterion for similarity.

So,

$BF/BE\text{ }=\text{ }BD/BC$

$BF/BE\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}$

[Since, D is the mid-point of BC]

$BE\text{ }=\text{ }2BF$

$BF\text{ }=\text{ }FE\text{ }=\text{ }2BF$

Thus,

$EF\text{ }=\text{ }FB$

(ii) In

$\vartriangle AFD,\text{ }EG\text{ }||\text{ }FD$

and using BPT we have

$AE/EF\text{ }=\text{ }AG/GD\text{ }\ldots .\text{ }\left( 1 \right)$

Now,

$AE\text{ }=\text{ }EB$

[Since, E is the mid-point of AB]

$AE\text{ }=\text{ }2EF\text{ }\left[ As,\text{ }EF\text{ }=\text{ }FB,\text{ }by\text{ }\left( 1 \right) \right]$

So, from (1) we have

$AG/GD\text{ }=\text{ }2/1$

Therefore,

$AG:\text{ }GD\text{ }=\text{ }2:\text{ }1$

i

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