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
In a ΔABC, E and F are the mid-points of AC and AB respectively. The altitude AP to BC
intersects FE at Q. Prove that AQ = QP.
In Fig. below, triangle ABC is right-angled at B. Given that AB = 9 cm, AC = 15 cm and D,
E are the mid-points of the sides AB and AC respectively, calculate
(i) The length of BC (ii) The area of ΔADE.
BM and CN are perpendiculars to a line passing through the vertex A of a triangle ABC. If
L is the mid-point of BC, prove that LM = LN.
In the given figure, seg PD is a median of ΔPQR. Point T is the mid point of seg PD. Produced QT intersects PR at M. Show that `"PM"/"PR" = 1/3`.
[Hint: DN || QM]
Prove that the figure obtained by joining the mid-points of the adjacent sides of a rectangle is a rhombus.
D, E, and F are the mid-points of the sides AB, BC and CA of an isosceles ΔABC in which AB = BC.
Prove that ΔDEF is also isosceles.
Prove that the figure obtained by joining the mid-points of the adjacent sides of a rectangle is a rhombus.
If L and M are the mid-points of AB, and DC respectively of parallelogram ABCD. Prove that segment DL and BM trisect diagonal AC.
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
In ΔABC, P is the mid-point of BC. A line through P and parallel to CA meets AB at point Q, and a line through Q and parallel to BC meets median AP at point R. Prove that: BC = 4QR
