Advertisements
Advertisements
Question
If P, Q and R are the mid-points of the sides BC, CA and AB of a triangle and AD is the perpendicular from A on BC, prove that P, Q, R and D are concyclic.
Advertisements
Solution

Given: In ΔABC, P, Q and R are the mid-points of the sides BC, CA and AB respectively. Also, AD ⊥ BC.
To prove: P, Q, R and D are concyclic.
Construction: Join DR, RQ and QP
Proof: In right-angled ΔADP, R is the mid-point of AB.
∴ RB = RD
⇒ ∠2 = ∠1 ...(i) [Angles opposite to the equal sides are equal]
Since, R and Q are the mid-points of AB and AC, then
RQ || BC ...[By mid-point theorem]
or RQ || BP
Since, QP || RB, then quadrilateral BPQR is a parallelogram.
⇒ ∠1 = ∠3 ...(ii) [Opposite angles of parallelogram are equal]
From equations (i) and (ii),
∠2 = ∠3
But ∠2 + ∠4 = 180° ...[Linear pair axiom]
∴ ∠3 + ∠4 = 180° ...[∴ ∠2 = ∠3]
Hence, quadrilateral PQRD is a cyclic quadrilateral.
So, points P, Q, R and D are non-cyclic.
Hence proved.
APPEARS IN
RELATED QUESTIONS
Prove that ‘Opposite angles of a cyclic quadrilateral are supplementary’.
ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD.
Prove that a cyclic parallelogram is a rectangle.
Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.
AC and BD are chords of a circle which bisect each other. Prove that (i) AC and BD are diameters; (ii) ABCD is a rectangle.
In a cyclic quadrilateral ABCD, if ∠A − ∠C = 60°, prove that the smaller of two is 60°
In the given figure, ABCD is a cyclic quadrilateral. Find the value of x.

ABCD is a cyclic quadrilateral in which BA and CD when produced meet in E and EA = ED. Prove that AD || BC .
In the figure, ▢ABCD is a cyclic quadrilateral. If m(arc ABC) = 230°, then find ∠ABC, ∠CDA, ∠CBE.

The three angles of a quadrilateral are 100°, 60°, 70°. Find the fourth angle.
