Advertisements
Advertisements
Question
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: AP = 2AR
Advertisements
Solution

In ΔABC,
P is the mid-point of BC and PQ is parallel to AC
Therefore, Q is the mid-point of AB.
In ΔABP,
Q is the mid-point of AB and QR is parallel to BP
Therefore, R is the mid-point of AP.
AR = RP
But AR + RP = AP
⇒ AR + AR = AP
⇒ 2AR = AP or AP = 2AR.
APPEARS IN
RELATED QUESTIONS
Fill in the blank to make the following statement correct
The triangle formed by joining the mid-points of the sides of an isosceles triangle is
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.

The figure, given below, shows a trapezium ABCD. M and N are the mid-point of the non-parallel sides AD and BC respectively. Find:

- MN, if AB = 11 cm and DC = 8 cm.
- AB, if DC = 20 cm and MN = 27 cm.
- DC, if MN = 15 cm and AB = 23 cm.
D and F are midpoints of sides AB and AC of a triangle ABC. A line through F and parallel to AB meets BC at point E.
- Prove that BDFE is a parallelogram
- Find AB, if EF = 4.8 cm.
A parallelogram ABCD has P the mid-point of Dc and Q a point of Ac such that
CQ = `[1]/[4]`AC. PQ produced meets BC at R.

Prove that
(i)R is the midpoint of BC
(ii) PR = `[1]/[2]` DB
In triangle ABC, D and E are points on side AB such that AD = DE = EB. Through D and E, lines are drawn parallel to BC which meet side AC at points F and G respectively. Through F and G, lines are drawn parallel to AB which meets side BC at points M and N respectively. Prove that: BM = MN = NC.
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.
ABCD is a kite in which BC = CD, AB = AD. E, F and G are the mid-points of CD, BC and AB respectively. Prove that: ∠EFG = 90°
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.
In the given figure, PS = 3RS. M is the midpoint of QR. If TR || MN || QP, then prove that:
RT = `(1)/(3)"PQ"`
