From above given triangle , It is clear that the altitudes AL, BM and CN of ?ABC intersect at H. Then H is the orthocentre of ?ABC.
In ?ABC, HL ? BC and BN ? CH.
Thus, the two altitudes HL and BN of ?HBC, intersects at A.
Hence the orthocentre of ?HBC is M .
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.
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
As per given figure , we can see that
Since the diagonals of a rectangle are equal and bisect each other. Let AC and BD intersect at M. Therefore M is the mid-point of AC and BD and AM = DM
From ?AOC, OA2 + OC2 = 2(AM2 + MO2) [Appollonius Theorem.] .......... ( 1 )
also in ?ODB, OB2 + OD2 = 2(MO2 + DM2) = 2(MO2 + AM2) .......... ( 2 )
From equations ( 1 ) and ( 2 ) .
? OA2 + OC2 = OB2 + OD2.
From above figure , we have
Since every exterior angle is equal to sum of interior opposite angles,
So, ? a = A + B, ? b = B + C and ? c = A + C
We know that , A + B + C = 180°
? ?a +?b + ?c = 2(A + B + C) = 2 × 180° = 360°.
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