Question

ABCD is a kite and∆A=∆C .If∆CAD=60°and∆CBD=45°; find
and angle∆BCD and∆CDA​

Answer 1

Answer:

∠BCD = 105°

∠CDA = 60°

Step-by-step explanation:

$\diamond$ Consider the point of intersection of the two diagonals as O

$\diamond$ In ΔAOD and ΔCOD,

$\implies$ AD = CD (Adjacent sides of kite are equal)

$\implies$ AO = CO (BD bisects AC)

$\implies$ OD = OD (Common)

∴, ΔAOD ≅ ΔCOD

$\diamond$ Hence,

∠CAD = ∠ACD = 60°

$\diamond$ Using angle sum property in ΔACD,

$\implies$ ∠CAD + ∠ ACD + ∠CDA = 180°

$\implies$ 60 + 60 + ∠CDA = 180

$\implies$ ∠CDA = 180 - 120

$\implies$ ∠CDA = 60°

$\diamond$ In ΔAOB and ΔCOB,

$\implies$ AO = CO (BD bisects AC)

$\implies$ OB = OB (Common)

$\implies$ AB = BC (Adjacent sides of kite are equal)

∴, ΔAOB ≅ ΔCOB

$\diamond$ Hence,

∠AOB = ∠COB

∠BAO = ∠BCO

$\diamond$ Using linear pair property,

$\implies$ ∠AOB + ∠COB = 180°

$\implies$ 2 * ∠AOB = 180

$\implies$ ∠AOB = 90° = ∠COB

$\diamond$ Using angle sum property in ΔBOC

$\implies$ ∠OBC + ∠COB + ∠BCO = 180

$\implies$ 45 + 90 + ∠BCO = 180

$\implies$ ∠BCO = 180 - 135

$\implies$ ∠BCO = 45°

$\diamond$ Therefore,

$\implies$ ∠BCD = ∠BCO + ∠DCO

$\implies$ ∠BCD = 45 + 60

$\implies$ ∠BCD = 105°

Answer 2

Answer:

angel cda=60° angle bcd= 105°