Question

In ΔABC, a line parallel to BC, passes through the mid-point of AB. Prove that the line bisects AC.

Answer 1

Step-by-step explanation:

HERE,

AD=BD (GIVEN)

DE II BC

CONSTRUCTION,

LET US EXTEND DE TO F AND JOIN F AND C IN SUCH A WAY THAT ABIICF

SO, BDFC IS A PARALLELOGRAM.

PROOF,

SINCE BDFC IS A PARALLELOGRAM

THEREFORE DBIIFC AND DB=FC

BUT AD=DB(GIVEN)

SO, AD=FC( EQUATION 1)

NOW,

IN TRIANGLE ADE AND CFE, WE HAVE,

ANGLE AED= ANGLE CEF(V.O.A)

ANGLE DAE= ANGLE FCE( ALTERNATE INTERIOR ANGLE)

AND AD=FC( FROM EQN 1)

THEREFORE, DE=FE (CPCT)

HENCE PROVED THAT E BISECTS AC TOO..

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Answer 2

Answer:

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