Question

P is the mid-point of side BC of a
parallelogram ABCD such that ZBAP =
ZDAP. If AD = 10 cm, then CD =​

Answer 1

GIVEN :-

• quadrilateral ABCD is a parallelogram

• hence , AB = CD , AB ll CD , BC = AD , BC ll AD

• P is the mid point of BC

• HENCE , BP = 1/2 × BC

• AD = BC = 10 cm

• angle BAP = angle DAP

TO FIND :-

• length of CD

SOLUTION :-

$\implies \rm{ let \: \angle apb = \angle1 }$

$\implies \rm{ now\: \angle dap = \angle1 } \:$

( alternative interior angles as AB ll DC )

$\implies \rm{ and \:also \: \angle bap = \angle dap \: \: \: \: \: \: (given) }$

hence ,

$\implies \rm{ \bf \angle bap = \angle1 }$

$\implies \rm{now \: if \: we \: see \: in \: \triangle apb}$

$\implies \rm{ \angle bap = \angle1 \: \: \: \: \: \: (proved \: above)}$

hence ∆APB is a isosceles triangle

so , opposite side will be equal

=> AB = BP eq (1)

now it's given that :-

$\implies \rm{bp = \dfrac{1}{2} \times bc}$

but BP = AB ( from equation 1 )

$\implies \rm{ab = \dfrac{1}{2} \times bc}$

now AD = BC

( given as AD ll BC And property of a parallelogram)

$\implies \rm{ab = \dfrac{1}{2} \times ad}$

now AD = 10 cm

$\implies \rm{ab = \dfrac{1}{2} \times 10}$

$\implies \rm{ \bf \: ab = 5cm}$

now AB = CD ( property of a parallelogram )

hence ,

$\implies \boxed{ \boxed{\rm{cd = 5cm}}}$