Quadrilaterals
Exercise 8.2 Part 2
Question 4: In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively. Show that the line segments AF and EC trisect the diagonal BD.
Answer: In ΔADE and ΔCBF
AD = BC (Opposite sides of parallelogram)
BF = DE (Half of opposite sides of parallelogram)
∠ADE = ∠CBF (Opposite angles)
So, ΔADE ≅ ΔCBF
Hence, AE = CF
In quadrilateral AECF
EC || AF & EC = AF
AE = CF
So, AE || CF
So, AECF is a parallelogram.
In Δ DQC
PE || QC (proved earlier by proving AE || CF)
E is the mid point of DC
So, P is the mid point of DQ
So, DP = PQ
In Δ APB
FQ || AP
F is the mid point of AB
So, PQ = QB
So, DP = PQ = QB proved
Question 5: Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.
Answer: ABCD is a quadrilateral in which P, Q, R, & S are mid points of AB, BC, CD & AD
In Δ ACD
SR is touching mid points of CD and AD
So, SR || AC
Similarly following can be proved
PQ || AC
QR || BD
PS || BD
So, PQRS is a parallelogram.
PR and QS are diagonals of the parallelogram PQRS, so they will bisect each other.
Question 6: ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that
- D is the midpoint of AC
- MD ⊥ AC
- `CM=MA=1/2\AB`
Answer: DM || BC
M is the mid point of AB
So, D is the mid point of AC (Mid point theorem)
`∠ACD=∠MDA=90°` (alternate angle to transversal MD)
Now in ΔCDM and ΔADM
`CD=AD`
`MD=MD`
`∠MDC=∠MDA`
So, `ΔCDM≅ ΔADM` (SAS theorem)
So, `MC=MA`
`MA=1/2\AB`
So, `MC=MA=1/2\AB`