Question

ABCD is a square. A is joined to point P on BC and D is joined to a point Q on AB. If AP=DQ, prove that AP⊥DQ.

Answer 1

DRP=90 Degree

Given DQ = AP

ABCD is a square.

To Find ∠DRP

In triangle APB and AQD

AP = DQ (given)

AB = AD (side of square)

R. H. S rule

Triangle ABP Triangle ADR

Hence

x = ∠ADQ = ∠PAB and

∠APB =∠AQD= y

Now

x + y = 900

In triangle APQ,

∠ARQ = 90 = 1800 (x + y) = 180 – 90 = 90

The proof is explained step-wise below :

Step-by-step explanation:

For better understanding of the solution, see the attached figure of the problem :

In ΔADQ and ΔAPB

AD = AB ( Sides of square are equal)

∠A = ∠B ( Each angle of square is right angle)

DQ = AP (Given)

So, By SAS congruency criterion, ΔADQ ≅ ΔAPB

⇒ ∠1 = ∠2 (Corresponding parts of the congruent triangles are equal)

Now, using angles sum property of triangle in ΔADM

∠1 + ∠3 + ∠4 = 180

⇒ ∠2 + ∠3 + ∠4 = 180 ( Because ∠1 = ∠2 )

But, ∠2 + ∠3 = 90

⇒ 90 + ∠4 = 180

⇒ ∠4 = 90°

Therefore, AP ⊥ DQ

Hence Proved.

Answer 2

Given: A square ABCD with AP = DQ

To prove: AP⊥DQ

Proof:

In Δs ADQ and APB,

∠A=∠B (Angles of a square is always 90°)

AD = AB (sides of a square)

DQ = AP (Given)

⇒ ΔADQ ≅ Δ APB (R.H.S. axiom)

⇒ ∠ADQ = ∠BAP (cpct).....(i)

Now in ΔADQ,

∠DAQ= 90°

⇒ ∠DAP + ∠PAB = 90°

⇒ ∠DAP + ∠ADQ = 90°....(ii)

i.e., ∠DAO +∠ADO =90° (from i)

∴ In Δ AOD,

∠DAO + ∠ADO+∠DOA = 180°

⇒ ∠DOA = 180°-90°

               = 90°

∴ DQ⊥AP (is proved).