The circumcentre of a triangle is always the point of intersection of the Bisectors .Hence required answer will be bisectors .
From given figure , we have
Let, (x + 1) be the hypotenuse and perpendicular = x - 1 , base = x.
By pythagoras theorem ,
? (x + 1)2 = x2 + (x + 1)2 ? x = 4
? Hypotenuse = 4 + 1 = 5
In a triangle ABC,
Given :- AB = 3 cm , AC = 5 cm and BC = 6 cm
We know that BD : DC = AB : AC
BD : DC = 3 : 5
? Divided BC = 6 in the ratio 3 : 5
BC = 3x + 5x = 6 ? 8x = 6 ? x = 6/8 = 0.75
? BD = 2.25, CD = 3.75.
We know that when two circle neither touch nor intersect and one lies outside the other, then 4 common tangents can be drawn.
For 4 common tangents , we get
? r1 + r2 < c1c2
Hence ,The number of tangents = 4
From above given figure ,
Proceeding as in Q. No. 4, we can prove that AXCY is a parallelogram .
Similarly, BXDY is a parallelogram.
Now, AXCY is a parallelogram .
? AY || CX
[? Opposite sides of a parallelogram are parallel]
? PY || QX ?(1)
Also, BXDY is a parallelogram
? DX || BY [? Opposite sides of a parallelogram are parallel]
? PX || QY ?(2)
Thus, in a quadrilateral PXQY,
From (i) and (ii) ,
we have , PY || QX and PX || QY
? PXQY is a parallelogram.
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