Question

ABCD is a cyclic quadrilateral. The diagonal BD bisects the diagonal AC, Show that
AB.AD = CB.CD.​

Answer 1

Step-by-step explanation:



GIVEN: ABCD is a cyclic quadrilateral. As it is inscribed in a circle. CD = CP , So, < D = < P = Z

CD = DQ, So, < DQC = < DCQ = X

< CAD = Y

TO FIND: < Y = ?

In triangle DAC

< Y = 180 - (Z + X) by angle sum property of triangle. …………(1)

< CAB = < CDB = 180 - 2X …………(2)

< ABP = Z , as exterior angle of a cyclic quadrilateral = interior opposite angle.

=> < PAB = 180 - 2Z ………….(3)

Now, by adding (1)+(2)+(3)

Y + < CAB + < BAP

= 180 - Z - X + 180–2X + 180 - 2Z = 180

=> 360 - 3Z - 3X = 0

=> 3X + 3Z = 360

=> X + Z = 120

Now, in triangle DCA , X + Z = 120

So, third angle Y = 180 - 120 = 60°

Ans : < CAD = 60°