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Question
Prove that the diagonals of a kite intersect each other at right angles.
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.
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