Advertisements
Advertisements
प्रश्न
In the given figure; ABC, AEQ and CEP are straight lines. Show that ∠APE and ∠CQE are supplementary.
उत्तर
Given – In the figure, ABC, AEQ and CEP are straight lines
To prove – ∠APE + ∠CQE = 180°
Construction – Join EB
Proof – In cyclic quad ABEP,
∠APE + ∠ABE = 180° ...(1)
Similarly, in cyclic quad BCQE
∠CQE + ∠CBE = 180° ...(2)
Adding (1) and (2),
∠APE + ∠ABE + ∠CQE + ∠CBE = 180° + 180° = 360°
`=>` ∠APE + ∠CQE + ∠ABE + ∠CBE = 360°
But, ∠ABE + ∠CBE = 180° ...[Linear pair]
∴ ∠APE + ∠CQE + 180° = 360°
`=>` ∠APE + ∠CQE = 360° – 180° = 180°
Hence ∠APE and ∠CQE are supplementary.
APPEARS IN
संबंधित प्रश्न
Two circle touch each other internally. Show that the tangents drawn to the two circles from any point on the common tangent are equal in length.
Two parallel tangents of a circle meet a third tangent at points P and Q. Prove that PQ subtends a right angle at the centre.
Two circles touch each other internally at a point P. A chord AB of the bigger circle intersects the other circle in C and D. Prove that ∠CPA = ∠DPB.
Two circles intersect at P and Q. Through P, a straight line APB is drawn to meet the circles in A and B. Through Q, a straight line is drawn to meet the circles at C and D. Prove that AC is parallel to BD.
Two circles intersect each other at points A and B. A straight line PAQ cuts the circles at P and Q. If the tangents at P and Q intersect at point T; show that the points P, B, Q and T are concyclic.
In the figure, given below, AC is a transverse common tangent to two circles with centres P and Q and of radii 6 cm and 3 cm respectively.
Given that AB = 8 cm, calculate PQ.
Two circles intersect each other at points C and D. Their common tangent AB touches the circles at point A and B. Prove that :
∠ ADB + ∠ ACB = 180°
Two circles with centres O and P intersect each other at A and B as shown in following fig. Two straight lines MAN and RBQ are drawn parallel to OP.
Prove that (i) MN = 20 P (ii) MN= RQ.
Two circles of radii 5cm and 3cm with centres O and P touch each other internally. If the perpendicular bisector of the line segment OP meets the circumference of the larger circle at A and B, find the length of AB.
Two circles with centres O and O' touch each other at point L. Prove that, a tangent through L bisects the common tangent AB of the two circles at point M.
Given: AB is a common tangent of the two circles that touch each other at point L. ML is a tangent through point L.
To prove: M is a mid-point of the tangent AB or MA = MB.
Proof: From the figure,
M is an external point that draws two tangents, MA and ML to the circle with the centre O.
So, `square` = `square` ......(i)
Similarly, M draws two tangents ML and MB to the circle with the centre O'.
So, `square` = `square` ......(ii)
From the equations (i) and (ii),
`square` = `square`
Hence, the tangent at the point L, bisects the common tangent, AB of the two circles at point M.
