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Standard X

Mathematics

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In the given ...

Question

$In the given figure, ABCD is a quadrilateral in which A B | | D C and P is the midpoint of BC. On producing, AP and DC meet at Q. Prove that (i) AB = CQ, (ii) DQ = DC + AB.$

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Solution

ANSWER:
In △ABP and △QCP, we have:

∠BPA=∠CPQ (Vertically opposite angle)

∠PAB=∠PQC
△ABP≅△QCP (AA criterion)
∴ AB = CQ
DQ=DC+CQ

⇒DQ=DC+AB
Hence, proved.

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Similar questions

In the adjoining figure, ABCD is a trapezium in which AB || DC and P, Q are the midpoints of AD and BC respectively. DQ and AB when produced meet at E. Also, AC and PQ A intersect at R. Prove that (i) DQ = QE, (ii) PR II AB and (iii) AR = RC.

Q. In the adjoining figure,

A B C D

is a trapezium in which

A B ∥ D C

and

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are the midpoint of

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respectively,

D Q

and

A B

when produced meet at

E

. Also ,

A C

and

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intersect at

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D Q = 2 Q E

Q. In the adjoining figure, ABCD is a trapezium in which AB || DC and P, Q are the midpoints of AD and BC respectively. DQ and AB when produced meet at E. Also, AC and PQ intersect at R. Prove that (i) DQ = QE, (ii) PR || AB, (iii) AR = RC.

Q. Question 8
In trapezium ABCD, AB || DC and L is the midpoint of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q. prove that ar (ABCD) = ar (APQD).

Q. In a trapezium ABCD, AB || DC and M is the midpoint of BC. Through M, a line PQ || AD has been drawn which meets AB in P and DC produced in Q, as shown in the adjoining figure. Prove that ar(ABCD) = ar(APQD).

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In the given figure, ABCD is a quadrilateral in which AB||DC and P is the midpoint of BC.On producing, AP and DC meet at Q. Prove that (i) AB = CQ, (ii) DQ = DC + AB.

ANSWER: In △ABP and △QCP, we have: ∠BPA=∠CPQ (Vertically opposite angle) ∠PAB=∠PQC △ABP≅△QCP (AA criterion) ∴ AB = CQ DQ=DC+CQ ⇒DQ=DC+AB Hence, proved.

$In the given figure, ABCD is a quadrilateral in which A B | | D C and P is the midpoint of BC. On producing, AP and DC meet at Q. Prove that (i) AB = CQ, (ii) DQ = DC + AB.$

ANSWER:
In △ABP and △QCP, we have:

∠BPA=∠CPQ (Vertically opposite angle)

∠PAB=∠PQC
△ABP≅△QCP (AA criterion)
∴ AB = CQ
DQ=DC+CQ

⇒DQ=DC+AB
Hence, proved.