Advertisements
Advertisements
प्रश्न
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
Advertisements
उत्तर
For help, we draw the diagonal BD as shown below
The diagonal AC and BD cuts at point X.
We know that the diagonal of a parallelogram intersect equally with each other. Therefore
AX = CX and BX = DX
Given,
CQ = `[1]/[4]`AC
CQ = `[1]/[4]` x 2CX
CQ = `[1]/[2]`CX
Therefore Q is the midpoint of CX.
(i) For triangle CDX PQ || DX or PR || BD
Since for triangle CBX
Q is the midpoint of CX and QR || BX. Therefore R is the midpoint of BC
(ii) For triangle BCD
As P and R are the mid-point of CD and BC, therefore PR = `[1]/[2]` DB
APPEARS IN
संबंधित प्रश्न
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.
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
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
The diagonals of a quadrilateral intersect at right angles. Prove that the figure obtained by joining the mid-points of the adjacent sides of the quadrilateral is rectangle.
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 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 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.
If L and M are the mid-points of AB, and DC respectively of parallelogram ABCD. Prove that segment DL and BM trisect diagonal AC.
In the given figure, ABCD is a trapezium. P and Q are the midpoints of non-parallel side AD and BC respectively. Find: DC, if AB = 20 cm and PQ = 14 cm
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.
