हिंदी

In a δAbc, E and F Are the Mid-points of Ac and Ab Respectively. the Altitude Ap to Bc Intersects Fe at Q. Prove that Aq = Qp.

Advertisements
Advertisements

प्रश्न

In a ΔABC, E and F are the mid-points of AC and AB respectively. The altitude AP to BC
intersects FE at Q. Prove that AQ = QP.

Advertisements

उत्तर

In ΔABC

E and F are midpoints of AB and  AC

∴ EF || FE, `1/2` BC =FE                [  ∴ By mid-point theorem]

In ΔABP

F is the midpoint of AB and   FQ || BP         [ ∵ EF || BC ]

∴ Q is the midpoint of AP                     [By converse of midpoint theorem]

Hence, AQ = QP

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 13: Quadrilaterals - Exercise 13.4 [पृष्ठ ६३]

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 9
अध्याय 13 Quadrilaterals
Exercise 13.4 | Q 5 | पृष्ठ ६३

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

In Fig. below, triangle ABC is right-angled at B. Given that AB = 9 cm, AC = 15 cm and D,
E are the mid-points of the sides AB and AC respectively, calculate
(i) The length of BC (ii) The area of ΔADE.

 


In Fig. below, BE ⊥ AC. AD is any line from A to BC intersecting BE in H. P, Q and R are
respectively the mid-points of AH, AB and BC. Prove that ∠PQR = 90°.


In the below Fig, ABCD and PQRC are rectangles and Q is the mid-point of Prove thaT

i) DP = PC (ii) PR = `1/2` AC


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.


D, E, and F are the mid-points of the sides AB, BC, and CA respectively of ΔABC. AE meets DF at O. P and Q are the mid-points of OB and OC respectively. Prove that DPQF is a parallelogram.


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.

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.


In a parallelogram ABCD, E and F are the midpoints of the sides AB and CD respectively. The line segments AF and BF meet the line segments DE and CE at points G and H respectively Prove that: ΔGEA ≅ ΔGFD


In AABC, D and E are two points on the side AB such that AD = DE = EB. Through D and E, lines are drawn parallel to BC which meet the side AC at points F and G respectively. Through F and G, lines are drawn parallel to AB which meet the side BC at points M and N respectively. Prove that BM = MN = NC.


In the given figure, PS = 3RS. M is the midpoint of QR. If TR || MN || QP, then prove that:

RT = `(1)/(3)"PQ"`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×