Question

in the figure abcd is a trapesium and ab parallel to cd. p is the mid point of ad if pq parallel to ab, then prove that q is the
midpoint of bc​

Answer 1

Step-by-step explanation:

nstruction: Join PB and extend it to meet CD produced at R.

To prove: PQ∥AB and PQ=

2

1

(AB+DC)

Proof : In △ABP and ΔDRP,

∠APB=∠DPR (Vertically opposite angles)

∠PDR=∠PAB (Alternate interior angles are equal)

AP=PD(P is the mid point of AD)

Thus, by ASA congruency,

ΔABP≅ΔDRP.

By CPCT,PB=PR and AB=RD

lnΔBRC

Q is the mid point of BC (Given)

P is the mid point of BR

(AsPB=PR)

So, by midpoint theorem, PQ ∥RC

⇒PQ∥DC

But AB∥DC(Given)

So,PQ∥AB

Also,PQ=

2

1

(RC)…. (using midpoint theorem)

PQ=

2

1

(RD+DC)

PQ=

2

1

(AB+DC)(∵AB=RD)

solution