Test: Mid Point Theorem - Grade 9 MCQ

Bcs (ABC) is a full triangle ABC triangle divide into four equal partsTherefore, 64÷4=16 cm²

Given that PQRS and ABCD are rectangle and Q is the mid pt of AC.

Since PQRS and ABCD are rectangle, every angle would be of 90°.

Consider AD and PQ.

angle ADP = 90°
angle QPC = 90°

Now, angle ADP lies corresponding to angle QPC and are equal.

If corresponding angles are equal, then the lines are parallel


=> AD || PQ

Now consider ∆ADC.

Q is the midpoint of AC (Given)
AD || PQ (shown above)

We know that,
line drawn parallel from the midpoint of a side, to other side, bisects the third side.

Hence, P would be the midpoint of DC

Hence, DP = PC

A median divides the triangle into 2 equal parts. So AD divides the triangle into the triangle ABD & ACD . Further BF divides the triangle into 2 equal parts .
So AF = 1/2 AC
DE bisects it into 2 parts
So EF = 1/2 EC
EF =FC
AF = EC
SO FC =1/2 * 1/2 AC
= 1/4 AC

Since, D,E,F are the midpoints of BC,CA and AB,Thus

Given,
ABC is a triangle.
D is midpoint of BC and DQ is drawn parallel to BA.
Then, Q is midpoint of AC.
∴ AQ = DC
∴ FA parallel to DQ||PC.
AQC, is a transversal so, AQ = QC and FDP also a transveral on them.
∴ FD = DP .......(1) [ intercept theorem]
EC = 1/2 AC = QC
Now, triangle EQD, here C is midpoint of EQ and CP which is parallel to DQ.
And, P is midpoint of DE.
DP = PE..........(2)
Therefore, From (1) and (2)
FD = DP = PE
∴ FD = 1/3 FE