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

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Answer 1

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

⇒ ∠ADQ = ∠BAP

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).

Answer 2

$\huge\blue{♡Answer♡}$

Given: A square ABCD with AP = DQ

To prove: AP | DQ

Prove:

In/\ ADQ and APB

/_ A = /_B Angles of a square is always 90•

AD = AB sides of square

DQ= AP (Given)

DQ = AP (proved)

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