ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (See the given figure). Which of the following should be true?

AB = BP

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$Δ A P B ≅ Δ C Q D$

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CQ = PQ

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Cannot be determined

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Solution

The correct option is B
$Δ A P B ≅ Δ C Q D$

$I n Δ A P B ≅ Δ C Q D$
$∠ A P B = ∠ C Q D (E a c h 90^{\circ})$
$AB = CD (Opposite sides of parallelogram ABCD)$
$∠ A B P = ∠ C D Q$ (Alternate interior angles for AB || CD)
$∴ Δ A P B ≅ Δ C Q D$ (By AAS congruency)

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Similar questions

ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (See the given figure). Then,

ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (See the given figure). Show that

(i) ΔAPB ≅ ΔCQD

(ii) AP = CQ

A B C D

A P

C Q

A

C

B D

A P = C Q

Δ A P B ≅ Δ C Q D

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