O is the point of intersection of the diagonals AC and BD of a trapezium ABCD with AB || DC. Through O, a line segment PQ is drawn parallel to AB meeting AD in P and BC in Q. Prove that PO = QO
Solution:
Given, ABCD is a trapezium with AB || DC.
O is the point of intersection of the diagonals AC and BD.
A line segment PQ is drawn parallel to AB through O.
P and Q intersect at AD and BC.
We have to prove that PO = QO.
In △ADC and △POA,
∠A = ∠A = common angle
The corresponding angles are equal. i.e.,∠ACD = ∠POA
AAA criterion states that if two angles of a triangle are respectively equal to two angles of another triangle, then by the angle sum property of a triangle their third angle will also be equal.
By AAA criterion, △ADC ⩬ △POA
By the property of similarity,
The corresponding sides are proportional.
So, PO/DC = AP/AD ------------------ (1)
In △BDC and △QOB,
∠B = ∠B = common angle
The corresponding angles are equal. i.e.,∠ACD = ∠QOB
By AAA criterion, △BDC ⩬ △QOB
By the property of similarity,
The corresponding sides are proportional.
So, OQ/DC = QB/BC -------------------- (2)
Now, in △ADB,
OP||AB
Basic Proportionality Theorem states that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
By basic proportionality theorem,
DP/AP = DO/OB --------------------- (3)
Now, in △CDB,
OQ||CD
By basic proportionality theorem,
CQ/QB = DO/OB --------------------- (4)
From (3) and (4),
DP/AP = CQ/QB
Adding 1 on both sides,
DP/AP + 1 = CQ/QB + 1
DP+AP / AP = CQ+QB / QB
From the figure,
DP + AP = AD
CQ + QB = BC
So, AD/AP = BC/QB
On rearranging,
AP/AD = QB/BC
From (1), PO/DC = QB/BC
From (2), PO/DC = OQ/DC
Therefore, it is clear that PO = OQ.
✦ Try This: In trapezium ABCD, AB || DC and L is the mid-point 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).
☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 6
NCERT Exemplar Class 10 Maths Exercise 6.4 Problem 15
O is the point of intersection of the diagonals AC and BD of a trapezium ABCD with AB || DC. Through O, a line segment PQ is drawn parallel to AB meeting AD in P and BC in Q. Prove that PO = QO
Summary:
O is the point of intersection of the diagonals AC and BD of a trapezium ABCD with AB || DC. Through O, a line segment PQ is drawn parallel to AB meeting AD in P and BC in Q. It is proven that PO = QO
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