Question

 □ABCD is a teapazium. AB || DC. Points P and Q are midpoints of seg AD and seg BC respectively.
Then prove that, PQ || AB and PQ
$\frac{1}{2}$(AB + DC).
​

Answer 1

Answer:

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

To prove: PQ || AB and PQ = 12 (AB + DC)

Proof : In ΔABP 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, Δ ABP ≅ Δ DRP.

By CPCT, PB = PR and AB = RD.

In Δ BRC,

Q is the mid point of BC (Given)

P is the mid point of BR (As PB = PR)

So, by midpoint theorem, PQ || RC

⇒ PQ || DC

But AB || DC (Given)

So, PQ || AB.

Also, PQ =

12

(RC) ....(using midpoint theorem)

PQ = 12 (RD + DC)

PQ =

AB + DC

12(AB + DC)

Step-by-step explanation:

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Answer 2

stop reporting revelent answers for points

specially mine _-_