P and Q are points on opposite sides AD and BC of a parallelogram ABCD such that PQ passes through the point of intersection O of its diagonals AC and BD. Show that PQ is bisected at O.
Solution:
Given, ABCD is a parallelogram
P and Q are the points on opposite sides AD and BC
The diagonals AC and BD intersect at O
PQ passes through the point of intersection O of its diagonals.
We have to show that PQ is bisected at O.
Considering triangles ODP and OBQ,
We know that the vertically opposite angles are equal.
∠BOQ = ∠POD
We know that the alternate interior angles are equal.
∠OBQ = ∠ODP
Given, OB = OD
ASA criterion states that two triangles are congruent if any two angles and the side included between them of one triangle are equal to the corresponding angles and the included side of the other triangle.
By ASA criteria, the triangles ODP and OBQ are congruent.
The Corresponding Parts of Congruent Triangles are Congruent (CPCTC) theorem states that when two triangles are congruent, then their corresponding sides and angles are also congruent or equal in measurements.
By CPCTC rule,
OP = OQ
Therefore, PQ is bisected at O.
✦ Try This: Can a quadrilateral ABCD be a parallelogram if ∠A = 70° and ∠C = 65°?
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 8
NCERT Exemplar Class 9 Maths Exercise 8.4 Problem 14
P and Q are points on opposite sides AD and BC of a parallelogram ABCD such that PQ passes through the point of intersection O of its diagonals AC and BD. Show that PQ is bisected at O.
Summary:
P and Q are points on opposite sides AD and BC of a parallelogram ABCD such that PQ passes through the point of intersection O of its diagonals AC and BD. It is shown that PQ is bisected at O
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