question
q: 4     in fig. , p is a point in the interior of a parallelogram abcd. show that

             (ii) 

             [hint : through p, draw a line parallel to ab.]
357
consider the parallelogram abcd with diagonals ac and bd as shown and they intersect at right angles at oa parallelogram is a square if its adjacent angles are 90° and adjacent sides are equalso we have to prove that adjacent sides of given parallelogram are equal and adjacent angles are 90° to prove given parallelogram is a squareconsider δbad and δcdaac = bd … givenab = dc … opposite sides of a parallelogramad is the common sidetherefore, δbad ≅ δcda … sss test for congruency⇒ ∠bad = ∠cda …corresponding angles of congruent triangles …(i)as it is given that abcd is parallelogram⇒ ∠bad + ∠cda = 180° … sum of adjacent angles of a parallelogram is 180°using equation (i)⇒ ∠cda + ∠cda = 180°⇒ 2 × ∠cda = 180°⇒ ∠cda = 90°⇒ ∠bad = 90°thus, adjacent angles are 90° … (iii)consider δaob and δaod∠aob = ∠aod … both 90° because given that diagonals intersect at right anglesod = ob … diagonals of a parallelogram bisect each otherao is the common sidetherefore, δaob ≅ δaod … sas test for congruency⇒ ab = ad … corresponding sides of congruent trianglesthus, adjacent sides are equal … (iv)from statements (iii) and (iv) we can conclude that parallelogram abcd is a squaretherefore, if in a parallelogram, the diagonals are equal in lengths and intersect at right angles, the parallelogram will be a square. given: ab=acto prove: ad>acproof:in ∆abc,∠acd = ∠b + ∠bac= ∠acb + ∠bac …as ∠c = ∠b as ab = ac= ∠cad + ∠cda +∠bac …as ∠acb = ∠cad + ∠cda∴∠acd > ∠cdaso the side opposite to ∠acd is the longest∴ad > ac ?solution 
step 1:
we have to draw the graph of form r to r. 
step 2:
the graph of f is the set of all points (x,f(x)) with x in the domain of f.
we take x in [-4,4]
and we will find the separate values of y for separate intervals.
 
 
 

 
......(1)
 

 

 

using x values along x-axis and along y-axis we get the graph of given function form r to r.
step 3:
the graph is as follow you seem to be missing some information about e.if e is the midpoint between d and f, thene is (1,1) take half of (4-2) = 1 and half of (5-3) = 1e alphabetically does come between d and f.just average the x values and average the y values of the two points d and for plot them graphically, carefully, maybe on some graph paper, and you'llvisually see the midpoint is at (1,1) if , passing through (1, 2), satisfies the differential equation , then we have to show that: a = b
given: p (a) = p (b)
let x be any element of set a,p(a) is the power set of set a, hence it contains all the subsets of set a. thus set a is contained in set p(a).

now let c be an element of set b
∴ 
now, we have:
c ⊂ b
∴ 
∴ a ⊂ b

similarly, we have:
b ⊂ a
now if a ⊂ b and b ⊂ a
∴ a = b a piece of paper is folded and punched as shown below the question figure. form the given figures, indicate how it will appear when opened.&nbsp;दर्शायी गई प्रश्न आकृति के अनुसार एक कागज के टुकड़े को मोड़ कर पंच किया जाता है, कागज़ खोलने पर किस उत्तर आकृति जैसा दिखाई देगा?&nbsp; (i) co-ordinates of point p are `((1xx17+2xx(-1))/(1+2),(1xx10+2xx1)/(1+2))` `=((17-8)/3,(10+2)/3)` `=(9/3,12/3)` `=(3,4)` (ii) `op=sqrt((0-3)^2+(0-4)^2)` `op=sqrt(9+16)` `op=sqrt25` `op=5` units (iii) let ab be divided by the point p(0, y) lying on y-axis in the ratio `therefore (0,y)=((kxx17+1xx(-4))/(k+1),(kxx10+1xx1)/(k+1))` `=>(0,y)=((17k-4)/(k+1),(10k+1)/(k+1))` `=>0=(17k-4)/(k+1)` `=>17k-4=0` `=>k=4/17` thus, the ratio in which the y-axis divide the line ab is 4: 17. May as well promote my comment to an answer.


You need to add points to a `Curve2D` object, then adjust the control points of the curve, and then finally draw the curve. So here's some simple, almost pseudo code:



```
var array_of_line_points # This already has the vectors which describe our line
for point in array_of_line_points:
  # The "get_perpendicular_vector()" function returns a vector that's a copy of the point, yet has been slid along a line parallel to two neighboring points. The "distance" variable is how far the control point should be from the originating point.
  var control_point1 = get_perpendicular_vector(point, distance)
  var control_point2 = get_perpendicular_vector(point, -distance)
  curve.add_point(point, control_point1, control_point2)
# ...In the draw function
func _draw():
  draw_polyline(curve.get_baked_points(), red, 2.0)

```

I got the inspiration for this from [Rob Spencer's page on Spline Interpolation](http://scaledinnovation.com/analytics/splines/aboutSplines.html). I'd like to make this more thorough (i.e. defining the `get_perpendicular_vector()` function). But this is the best I can currently do. in this we are asked to divide it in 2:3 ratioso, 3+2 = 55, number congruent of line segment is formed.i. firstly, we have to draw a line segment ab of length 7 cm. draw a line with an acute angle from point aii. point the compass on a, and cut the new line with some radiusplace the compass along the line and make 5 arcs. since we need 5 congruent parts, and name the last one as “g”iii. take length of bg as radius draw an arc from a just below it.again, take the length of ag as radius and draw an arc which cuts the previous arc from b, name the intersection point as hconnect b and hiv. taking the same length that we used in line ag, draw arcs along bh and name the points.v. connect the points along ag and bh. name the intercepts on the line ab as m, n, o & p.vi. thus we can divide a line segment as ao:ob = 6:4= 3:2