Question

In the quadrilateral PQRS, PQ is the smallest side and SR is the longest side. Prove that ∠P > ∠R and ∠Q > ∠S.

[Hint: Join PR, prove that x > y and a > b. ∴ x + a > y + b, etc.]

Answer 1

Given

In quadrilateral PQRS:

• PQ is the smallest side,
• SR is the longest side.

To prove: ∠P > ∠R and ∠Q > ∠S. 

1) Prove ∠P > ∠R

Join the diagonal PR.

• In ΔPQR, let ∠QPR = x and ∠PRQ = y.
  Since PQ is the smallest side, the angle opposite it is the smallest angle of ΔPQR.
  Opposite PQ is ∠PRQ = y.
  Hence x > y.
• In ΔPRS, let ∠RPS = a and ∠PRS = b.
  Since SR is the longest side, the angle opposite it is the largest angle of ΔPRS.
  Opposite SR is ∠RPS = a.
  Hence a > b.

Now the interior angles at P and R of the quadrilateral are:

∠P = ∠QPR + ∠RPS = x + a,

∠R = ∠PRQ + ∠PRS = y + b.

Since x > y and a > b, we get x + a > y + b.

Thus ∠P > ∠R.

2) Prove ∠Q > ∠S

Now join the other diagonal QS.

• In ΔPQS, denote ∠PQS = u and ∠QSP = v.
  Here PQ is still the smallest side, so the angle opposite PQ it is the smallest angle of ΔPQS.
  Opposite PQ is ∠QSP = v.
  Hence u > v.
• In ΔQRS, denote ∠RQS = m and ∠QSR = n.
  Here SR is the longest side, so the angle opposite SR is the largest angle of ΔQRS.
  Opposite SR is ∠RQS = m.
  Hence m > n.

The interior angles at Q and S of the quadrilateral are:

∠Q = ∠PQS + ∠RQS = u + m,

∠S = ∠QSP + ∠QSR = v + n.

Since u > v and m > n, we get u + m > v + n.

Thus, ∠Q > ∠S. 

Therefore, using the “greater side ↔ greater opposite angle” fact in the two triangles formed by each diagonal, we conclude:

∠P > ∠R and ∠Q > ∠S