Question

in the given figure, ABCD is a square and angle PQR = 90°. If PB = QC = DR, prove that
(i) QB = RC, (ii) PQ=OR, (iii) angle QPR = 45°
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Answer 1

Here's your answer!

(i) QB = RC

We know that the line segment QB can be written as QB = BC – QC

Since ABCD is a square we know that BC = DC and QC = DR

So we get QB = CD – DR

From the figure

we get QB = RC

(ii)PQ = OR

Consider △ PBQ and △ QCR

It is given that PB = QC

Since ABCD is a square we get

∠ PBQ = ∠ QCR = 90°

By SAS congruence criterion

△ PBQ ≅ △ QCR PQ = QR (c. p. c. t rule )

(iii) ∠QPR = 45°

We know that PQ = QR Consider

△ PQR

From the figure we know that ∠ QPR and ∠ QRP are base angles of isosceles triangle

∠ QPR = ∠ QRP

We know that the sum of all the angles in a triangle is 180°

∠ QPR + ∠ QRP + ∠ PRQ = 180°

By substituting the values in the above equation

∠ QPR + ∠ QRP + 90° = 180°

∠ QPR + ∠ QRP = 180° – 90°

∠ QPR + ∠ QRP = 90°

We know that, ∠ QPR = ∠ QRP

So we get, ∠ QPR + ∠ QPR = 90°

2∠ QPR = 90°

∠ QPR = 45°

∴ ∠ QPR = 45°.

The diagram :