Advertisements
Advertisements
Question
In the given figure, T is the midpoint of QR. Side PR of ΔPQR is extended to S such that R divides PS in the ratio 2:1. TV and WR are drawn parallel to PQ. Prove that T divides SU in the ratio 2:1 and WR = `(1)/(4)"PQ"`.
Advertisements
Solution
In ΔPQR,
T is the mid-point of QR and VT || PQ
So, V is the mid-point of PR.
Since R divides PR in the ratio 2 : 1 and PV = VR,
so, PV = PR = RS
Since R is the mid-point of SV and RW || VT,
W is the mid-point of ST.
Since V is the mid-point of PR and VT || PQ,
T is the mid-point of UW.
So, UT = TW = SW
⇒ T divides SU in the ratio 2 : 1
Also,
R and W are the midpoints SV and TS respectively.
⇒ WR = `(1)/(2)"VT"`
V and T are the mid-points of PR and UW respectively.
⇒ VT = `(1)/(2)"PQ"`
So, WR = `(1)/(2)(1/2 "PQ")`
⇒ WR = `(1)/(4)"PQ"`.
APPEARS IN
RELATED QUESTIONS
In below fig. ABCD is a parallelogram and E is the mid-point of side B If DE and AB when produced meet at F, prove that AF = 2AB.

In a ΔABC, BM and CN are perpendiculars from B and C respectively on any line passing
through A. If L is the mid-point of BC, prove that ML = NL.
In the Figure, `square`ABCD is a trapezium. AB || DC. Points P and Q are midpoints of seg AD and seg BC respectively. Then prove that, PQ || AB and PQ = `1/2 ("AB" + "DC")`.

In the adjacent figure, `square`ABCD is a trapezium AB || DC. Points M and N are midpoints of diagonal AC and DB respectively then prove that MN || AB.

In triangle ABC, AD is the median and DE, drawn parallel to side BA, meets AC at point E.
Show that BE is also a median.
In trapezium ABCD, AB is parallel to DC; P and Q are the mid-points of AD and BC respectively. BP produced meets CD produced at point E.
Prove that:
- Point P bisects BE,
- PQ is parallel to AB.
In triangle ABC; M is mid-point of AB, N is mid-point of AC and D is any point in base BC. Use the intercept Theorem to show that MN bisects AD.
Side AC of a ABC is produced to point E so that CE = `(1)/(2)"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 meets AC at point P and EF at point R respectively. Prove that: 4CR = AB.
In ΔABC, X is the mid-point of AB, and Y is the mid-point of AC. BY and CX are produced and meet the straight line through A parallel to BC at P and Q respectively. Prove AP = AQ.
The figure obtained by joining the mid-points of the sides of a rhombus, taken in order, is ______.
