Question

P is the centre of the circle. Chord AB and chord CD intersect on the diameter at the point Q. If $\angle$AQP $\cong$ $\angle$PQD, then AB = CD.

• A. True
• B. False

Answer 1

The correct option is A True


Draw PM$\perp$CD and PN$\perp$AB.
Now, in $△$PMQ and $△$PNQ, we have
$\angle$PMQ = $\angle$PNQ = 90${}^{\circ }$ [by construction]
$\angle$PQM $\cong$ $\angle$PQN [given]
PQ is the common side
By AAS criterion, $△$PMQ $\cong$ $△$PNQ
$⟹$ PM = PN [CPCT]
We know that the chords equidistant from the centre are equal.
Hence AB = CD.
Therefore the statement is true.