Advertisements
Advertisements
Question
Prove that the diagonals of a kite intersect each other at right angles.
Advertisements
Solution
Consider ABCD is a kite.
Then, AB = AD and BC = DC
In ΔABC and ΔADC,
AB = AD
BC = CD
AC = AC
∴ ΔABC ≅ ΔADC ...(SSS Congruence)
⇒ ∠BCA = ∠DCA ...(C.P.C.T)
⇒ ∠BCO = ∠DCO ...(i)
Now,
In ΔOBC and ΔODC,
BC = CD
∠BCO = ∠DCO ...[From (i)]
OC = OC ...(common)
∴ ΔOBC ≅ ΔODC ...(SAS congruence)
∠COB - ∠COD ...(C. C.T)
But, ∠COB + ∠CID = 180° ...(Linear pair)
⇒ ∠COB + ∠COB = 180°
⇒ 2∠COB = 180°
⇒ ∠COB = 90°
Hence, diagonals f a kite intersect each other at right angles.
APPEARS IN
RELATED QUESTIONS
SN and QM are perpendiculars to the diagonal PR of parallelogram PQRS.
Prove that:
(i) ΔSNR ≅ ΔQMP
(ii) SN = QM
Prove that if the diagonals of a parallelogram are equal then it is a rectangle.
P is a point on side KN of a parallelogram KLMN such that KP : PN is 1 : 2. Q is a point on side LM such that LQ : MQ is 2 : 1. Prove that KQMP is a parallelogram.
Prove that the line segment joining the mid-points of the diagonals of a trapezium is parallel to each of the parallel sides, and is equal to half the difference of these sides.
In a parallelogram PQRS, M and N are the midpoints of the opposite sides PQ and RS respectively. Prove that
RN and RM trisect QS.
In a parallelogram PQRS, M and N are the midpoints of the opposite sides PQ and RS respectively. Prove that
PMRN is a parallelogram.
In a parallelogram PQRS, M and N are the midpoints of the opposite sides PQ and RS respectively. Prove that
MN bisects QS.
The diagonals AC and BC of a quadrilateral ABCD intersect at O. Prove that if BO = OD, then areas of ΔABC an ΔADC area equal.
In the given figure, PQ ∥ SR ∥ MN, PS ∥ QM and SM ∥ PN. Prove that: ar. (SMNT) = ar. (PQRS).
In ΔPQR, PS is a median. T is the mid-point of SR and M is the mid-point of PT. Prove that: ΔPMR = `(1)/(8)Δ"PQR"`.
