Advertisements
Advertisements
प्रश्न
If the sides of a quadrilateral ABCD touch a circle, prove that : AB + CD = BC + AD.
उत्तर
Let the circle touch the sides AB, BC, CD and DA of quadrilateral ABCD at P, Q, R and S respectively.
Since AP and AS are tangents to the circle from external point A
AP = AS ...(i)
Similarly, we can prove that:
BP = BQ ...(ii)
CR = CQ ...(iii)
DR = DS ...(iv)
Adding,
AP + BP + CR + DR = AS + DS + BQ + CQ
AB + CD = AD + BC
Hence, AB + CD = AD + BC
APPEARS IN
संबंधित प्रश्न
In a triangle ABC, the incircle (centre O) touches BC, CA and AB at points P, Q and R respectively. Calculate :
- ∠QOR
- ∠QPR;
given that ∠A = 60°.
In the figure; PA is a tangent to the circle, PBC is secant and AD bisects angle BAC. Show that triangle PAD is an isosceles triangle. Also, show that:
`∠CAD = 1/2 (∠PBA - ∠PAB)`
In the given figure, O is the centre of the circle. Tangents at A and B meet at C. If ∠ACO = 30°, find:
- ∠BCO
- ∠AOB
- ∠APB
In the given figure, XY is the diameter of the circle and PQ is a tangent to the circle at Y.
If ∠AXB = 50° and ∠ABX = 70°, find ∠BAY and ∠APY.
In the given figure, PT touches a circle with centre O at R. Diameter SQ when produced to meet the tangent PT at P. If ∠SPR = x° and ∠QRP = y°; Show that x° + 2y° = 90°
In Fig. l and m are two parallel tangents at A and B. The tangent at C makes an intercept DE between n and m. Prove that ∠ DFE = 90°
A circle touches the sides of a quadrilateral ABCD at P, Q, R, S respectively. Show that the angles subtended at the centre by a pair of opposite sides are supplementary.
In Fig. AT is a tangent to the circle. If m∠ABC = 50°, AC = BC, Find ∠BAT.
In the given figure, AC is a tangent to circle at point B. ∆EFD is an equilateral triangle and ∠CBD = 40°. Find:
- ∠BFD
- ∠FBD
- ∠ABF
In the given figure PT is a tangent to the circle. Chord BA produced meets the tangent PT at P.
Given PT = 20 cm and PA = 16 cm.
- Prove ΔPTB ~ ΔPAT
- Find the length of AB.
