Chapter 15

Solution 12
It is given that ABCD is a kite in which ∠CAD=70°, ∠CBD=65° and ∠A=∠C.

Now, in ΔACD, we have

∠A+∠C+∠D=180° (Angle sum property)

⇒∠CAD+∠ADC+∠ACD=180°

⇒70°+70°+∠ADC=180°

⇒∠ADC=40°

Now, From ABCD, we have

∠A+∠B+∠C+∠D=360° (Sum of all the angles of quadrilateral is 360°)

⇒∠BAD+∠ABC+∠BCD+∠CDA=360°

⇒2(∠CAD)+2(∠CBD)+∠BCD+∠CDA=360°

⇒2(70)+2(65)+∠BCD+40=360°

⇒∠BCD=50°

Solution 13