Mathematics
ABCD is a square. A is joined to a point P on BC and D is joined to a point Q on AB. If AP = DQ, prove that AP and DQ are perpendicular to each other.
Answer
Square ABCD is shown in the figure below:
Considering △ABP and △ADQ we have,
⇒ ∠ABP = ∠DAQ = 90°
⇒ AP = DQ (Given)
⇒ AB = AD (Sides of square are equal)
Hence, △ABP ≅ △ADQ by RHS axiom.
⇒ ∠BAP = ∠ADQ (By C.P.C.T.) ……..(i)
⇒ ∠BAD = 90° (Each angle of square = 90°)
⇒ ∠BAP + ∠PAD = 90°
Substituting value of ∠ADQ from (i) we get,
⇒ ∠ADQ + ∠PAD = 90° ………(ii)
From figure,
∠ADQ = ∠ADM
∠PAD = ∠MAD
Substituting above values in (ii) we get,
⇒ ∠ADM + ∠MAD = 90° …….(iii)
In △AMD,
⇒ ∠ADM + ∠MAD + ∠AMD = 180°
⇒ 90° + ∠AMD = 180° (From iii)
⇒ ∠AMD = 90°
∴ AP ⊥ DQ.
Hence, proved that AP ⊥ DQ.
Related Questions
In figure (1) given below, ABCD is a parallelogram and X is mid-point of BC. The line AX produced meets DC produced at Q. The parallelogram ABPQ is completed. Prove that
(i) the triangles ABX and QCX are congruent.
(ii) DC = CQ = QP
In figure (2) given below, points P and Q have been taken on opposite sides AB and CD respectively of a parallelogram ABCD such that AP = CQ. Show that AC and PQ bisect each other.
If P and Q are points of trisection of the diagonal BD of a parallelogram ABCD, prove that CQ || AP.
A transversal cuts two parallel lines at A and B. The two interior angles at A are bisected and so are the two interior angles at B; the four bisectors form a quadrilateral ACBD. Prove that
(i) ACBD is a rectangle.
(ii) CD is parallel to the original parallel lines.