मराठी

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
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:

  1. M, A, and N are collinear.
  2. A is the mid-point of MN.
बेरीज
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.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 12: Mid-point and Its Converse [ Including Intercept Theorem] - Exercise 12 (B) [पृष्ठ १५४]

APPEARS IN

सेलिना Concise Mathematics [English] Class 9 ICSE
पाठ 12 Mid-point and Its Converse [ Including Intercept Theorem]
Exercise 12 (B) | Q 4 | पृष्ठ १५४

संबंधित प्रश्‍न

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:

  1. SR || AC and SR = `1/2AC`
  2. PQ = SR
  3. PQRS is a parallelogram.


ABCD is a rhombus. EABF is a straight line such that EA = AB = BF. Prove that ED and FC when produced, meet at right angles.


Fill in the blank to make the following statement correct:

The triangle formed by joining the mid-points of the sides of a right triangle is            


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, P is the mid-point of side 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 : (i) AP = 2AR
                   (ii) BC = 4QR


In the figure, give below, 2AD = AB, P is mid-point of AB, Q is mid-point of DR and PR // BS. Prove that:
(i) AQ // BS
(ii) DS = 3 Rs.


In ΔABC, D, E, F are the midpoints of BC, CA and AB respectively. Find DE, if AB = 8 cm


In the given figure, ABCD is a trapezium. P and Q are the midpoints of non-parallel side AD and BC respectively. Find: PQ, if AB = 12 cm and DC = 10 cm.


In ΔABC, D and E are the midpoints of the sides AB and BC respectively. F is any point on the side AC. Also, EF is parallel to AB. Prove that BFED is a parallelogram.

Remark: Figure is incorrect in Question


The quadrilateral formed by joining the mid-points of the sides of a quadrilateral PQRS, taken in order, is a rhombus, if ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×