Advertisements
Advertisements
प्रश्न
In triangle ABC, the medians BP and CQ are produced up to points M and N respectively such that BP = PM and CQ = QN. Prove that:
- M, A, and N are collinear.
- A is the mid-point of MN.
Advertisements
उत्तर
The figure is shown below

(i) In ΔAQN & ΔBQC
AQ = QB (Given)
∠AQN = ∠BQC
QN = QC
∴ ΔAQN ≅ ΔBQC ...[ by SAS ]
∴ ∠QAN = ∠QBC ...(1)
And BC = AN ……(2)
Similarly, ΔAPM ≅ ΔCPB .....[by SAS]
∠PAM = ∠PCB ...(3) [by CPTC]
And BC = AM ….( 4 )
Now In ΔABC,
∠ABC + ∠ACB + ∠BAC = 180°
∠QAN + ∠PAM + ∠BAC = 180° ...[ (1), (2) we get ]
Therefore M, A, N are collinear.
(ii) From (3) and (4) MA = NA
Hence A is the midpoint of MN.
APPEARS IN
संबंधित प्रश्न
In the given figure, ΔABC is an equilateral traingle. Points F, D and E are midpoints of side AB, side BC, side AC respectively. Show that ΔFED is an equilateral traingle.

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.
In ∆ABC, E is the mid-point of the median AD, and BE produced meets side AC at point Q.
Show that BE: EQ = 3: 1.
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, sides AB and DC are parallel to each other. E is mid-point of AD and F is mid-point of BC.
Prove that: AB + DC = 2EF.
In ΔABC, BE and CF are medians. P is a point on BE produced such that BE = EP and Q is a point on CF produced such that CF = FQ. Prove that: A is the mid-point of PQ.
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
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, D and E are the midpoints of the sides AB and AC respectively. F is any point on the side BC. If DE intersects AF at P show that DP = PE.
