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
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Answer:
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).
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.