1. Fig.9.23, E is any point on median AD of a
ABC. Show that ar (ABE) = ar (ACE).
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Given :- AD is a median of ABC.
To Prove :- ar.(ABE) = ar.(ACE)
Proof :-
In ABC,
AD is the median.
So, ar.(ABD) = ar.(ACD) -- (Equation 1)
In BEC,
ED is the median.
So, ar.(BED) = ar.(CED) -- (Equation 2)
Now, Equation 1 – Equation 2.
➣ ar.(ABD) – ar.(BED) = ar.(ACD) – ar.(CED)
➣ ar.(ABE) = ar.(ACE)
Hence, Proved.
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: Solution
Given,
➝ AD is median of ΔABC it will divide ΔABC into two triangles of equal area.
➝ ar(ABD) = ar(ACD) — (i)
also,
➢ ED is the median of ΔABC.
ar(EBD) = ar(ECD) — (ii)
➝ Subtracting (ii) from (i),
➢ ar(ABD) – ar(EBD) = ar(ACD) – ar(ECD)
➢ ar(ABE) = ar(ACE).
hence proved
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