मराठी

In Parallelogram Abcd, E and F Are Mid-points of the Sides Ab and Cd Respectively. the Line Segments Af and Bf Meet the Line Segments Ed and Ec at Points G and H Respectively - Mathematics

Advertisements
Advertisements

प्रश्न

In parallelogram ABCD, E and F are mid-points of the sides AB and CD respectively. The line segments AF and BF meet the line segments ED and EC at points G and H respectively.
Prove that:
(i) Triangles HEB and FHC are congruent;
(ii) GEHF is a parallelogram.

बेरीज
Advertisements

उत्तर

The figure is shown below

(i) From ΔHEB and ΔFHC
BE = FC
∠EHB = ∠FHC                        ...[ Opposite angle ]
∠HBE = ∠HFC
∴  ΔHEB ≅ ΔFHC
∴  EH = CH , BH = FH

(ii) Similarly AG = GF and EG = DG      …..(1)
For triangle ECD,
F and H are the mid-point of CD and EC.
Therefore HF || DE and 
HF = `[1]/[2]` DE                                          ....(2)

From (1) and (2) we get,
HF = EG and HF || EG
Similarly, we can show that EH = GF and EH || GF
Therefore GEHF is a parallelogram.

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 6 | पृष्ठ १५४

संबंधित प्रश्‍न

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.


ABC is a triang D is a point on AB such that AD = `1/4` AB and E is a point on AC such that AE = `1/4` AC. Prove that DE = `1/4` BC.


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: QAP is a straight line.


The diagonals of a quadrilateral intersect each other at right angle. Prove that the figure obtained by joining the mid-points of the adjacent sides of the quadrilateral is a rectangle.


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 ΔABC, the medians BE and CD are produced to the points P and Q respectively such that BE = EP and CD = DQ. Prove that: Q A and P are collinear.


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


The figure obtained by joining the mid-points of the sides of a rhombus, taken in order, is ______.


P and Q are the mid-points of the opposite sides AB and CD of a parallelogram ABCD. AQ intersects DP at S and BQ intersects CP at R. Show that PRQS is a parallelogram.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×