Advertisements
Advertisements
Question
The side AC of a triangle ABC is produced to point E so that CE = AC. D is the mid-point of BC and ED produced meets AB at F. Lines through D and C are drawn parallel to AB which meet AC at point P and EF at point R respectively.
Prove that:
- 3DF = EF
- 4CR = AB
Advertisements
Solution
Consider the figure :
Here D is the midpoint of BC and DP is parallel to AB, therefore P is the midpoint of AC and
PD = `[1]/[2]`AB
(i) Again from the triangle AEF, we have AF || PD || CR and AP = `[1]/[3]`AE
Therefore, DF = `[1]/[3]` EF or we can say that 3DF = EF.
Hence, it is shown.
(ii) From the triangle PED, we have PD || CR and C is the midpoint of PE, therefore, CR = `[1]/[2]`PD
Now,
PD = `1/2` AB
`1/2"PD" = 1/4`AB
CR = `1/4`AB
4CR = AB
Hence, it is shown.
APPEARS IN
RELATED QUESTIONS
ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see the given figure). AC is a diagonal. Show that:
- SR || AC and SR = `1/2AC`
- PQ = SR
- PQRS is a parallelogram.
ABC is a triangle and through A, B, C lines are drawn parallel to BC, CA and AB respectively
intersecting at P, Q and R. Prove that the perimeter of ΔPQR is double the perimeter of
ΔABC
In Fig. below, BE ⊥ AC. AD is any line from A to BC intersecting BE in H. P, Q and R are
respectively the mid-points of AH, AB and BC. Prove that ∠PQR = 90°.
In below Fig, ABCD is a parallelogram in which P is the mid-point of DC and Q is a point on AC such that CQ = `1/4` AC. If PQ produced meets BC at R, prove that R is a mid-point of BC.
In parallelogram ABCD, E and F are mid-points of the sides AB and CD respectively. The line segments AF and BF meet the line segments ED and EC at points G and H respectively.
Prove that:
(i) Triangles HEB and FHC are congruent;
(ii) GEHF is a parallelogram.
Prove that the figure obtained by joining the mid-points of the adjacent sides of a rectangle is a rhombus.
In the given figure, ABCD is a trapezium. P and Q are the midpoints of non-parallel side AD and BC respectively. Find: AB, if DC = 8 cm and PQ = 9.5 cm
ABCD is a parallelogram.E is the mid-point of CD and P is a point on AC such that PC = `(1)/(4)"AC"`. EP produced meets BC at F. Prove that: F is the mid-point of BC.
In ΔABC, the medians BE and CD are produced to the points P and Q respectively such that BE = EP and CD = DQ. Prove that: Q A and P are collinear.
Show that the quadrilateral formed by joining the mid-points of the consecutive sides of a square is also a square.
