Question

ABCD is a square and P is any point on BC. The perpendicular drawn from B on AP meets DC at Q. Prove that AP=BQ

Answer 1

Answer:

The figure according to given information is as shown:

Mark the intersection point of AP and BQ as O

(See fig. No. = A)

Consider ∠BQC be x as shown

Consider ΔBCQ

∠C = 90° … ABCD is a square

As sum of angles of a triangle is 180°

⇒ ∠C + ∠BQC + ∠QBC = 180°

⇒ 90° + x + ∠QBC = 180°

⇒ ∠QBC = 90° - x …(i)

Now consider ΔOBP

∠BOP = 90° … given AP perpendicular to BQ

∠OBP = 90° – x … using (i)

As sum of angles of a triangle is 180°

⇒ ∠BOP + ∠OBP + ∠OPB = 180°

⇒ 90° + 90° - x + ∠OPB = 180°

⇒ ∠OPB = x … (ii)

Consider ΔAPB

∠ABP = 90° … ABCD is a square

∠APB = x … using (ii)

As sum of angles of a triangle is 180°

⇒ ∠ABP + ∠APB + ∠BAP = 180°

⇒ 90° + x + ∠BAP = 180°

⇒ ∠BAP = 90° - x … (iii)

Consider ΔAPB and Δ BQC

These two triangles are drawn separately from the same figure given above

(See fig. No. = B)

These angles are written using (i), (ii) and (iii)

∠QCB = ∠ABP = 90° … angles of a square ABCD

BC = AB … sides of a square ABCD

∠QBC = ∠PAB = 90° - x …using (i) and (iii)

Therefore, ΔQCB ≅ ΔPBA … ASA test for congruency

⇒ BQ = AP … corresponding sides of congruent triangles

Hence proved.