In Δ ABC, E is the mid-point of median AD such that BE produced meets AC at F. If AC = 10.5 cm, then AF =
A. 3 cm
B. 3.5 cm
C. 2.5 cm
D. 5 cm
Answers
Given : In Δ ABC, E is the mid-point of median AD such that BE produced meets AC at F and AC = 10.5 cm.
To find : The value of AF
Construct : Draw DM || EF.
Proof :
In ∆ABC,
E is mid point of median AD
We have, AC = 10.5 cm
∵ E is mid point of AD so F is mid point of AM
[By converse of mid point theorem]
∴ AF = FM ……………………(1)
In ∆BFC,D is the mid point of BC
EF || DM
G is the mid point of CF.
[By converse of mid point theorem]
So, FM = MC ………………….(2)
From eq (1) & (2),
AF = FM = MC……………..(3)
AC = AF + MC + FM
AC = AF + AF + AF
[From eq (3) ]
AC = 3AF
AF = AC/3
AF = ⅓ × 10.5
AF = 3.5 cm
Hence , AF is 3.5 cm.
Among the given options option (B) 3.5 cm is correct.
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Answer:
Step-by-step explanation:
In ∆ABC,
E is mid point of median AD
We have, AC = 10.5 cm
∵ E is mid point of AD so F is mid point of AM
[By converse of mid point theorem]
∴ AF = FM ……………………(1)
In ∆BFC,D is the mid point of BC
EF || DM
G is the mid point of CF.
[By converse of mid point theorem]
So, FM = MC ………………….(2)
From eq (1) & (2),
AF = FM = MC……………..(3)
AC = AF + MC + FM
AC = AF + AF + AF
[From eq (3) ]
AC = 3AF
AF = AC/3
AF = ⅓ × 10.5
AF = 3.5 cm