Advertisements
Advertisements
प्रश्न
Through the mid-point M of the side CD of a parallelogram ABCD, the line BM is drawn intersecting diagonal AC in L and AD produced in E. Prove that: EL = 2BL.
उत्तर
∠1 = ∠6 ...(Alternate interior angles)
∠2 = ∠3 ...(Vertically opposite angles)
DM = MC ...(M is the mid-point of CD)
∴ ∆DEM ≅ ∆CBM ...(AAS congruence criterion)
So, DE = BC ...(Corresponding parts of congruent triangles)
Also, AD = BC ...(Opposite sides of a parallelogram)
`=>` AE = AD + DE = 2BC
Now, ∠1 = ∠6 and ∠4 = ∠5
∴ ∆ELA ~ ∆BLC ...(AA similarity)
`=> (EL)/(BL) = (EA)/(BC)`
`=> (EL)/(BL) = (2BC)/(BC) = 2`
`=>` EL = 2BL
संबंधित प्रश्न
Given : AB || DE and BC || EF. Prove that :
- `(AD)/(DG) = (CF)/(FG)`
- ∆DFG ∼ ∆ACG
In the given figure, P is a point on AB such that AP : PB = 4 : 3. PQ is parallel to AC.
- Calculate the ratio PQ : AC, giving reason for your answer.
- In triangle ARC, ∠ARC = 90° and in triangle PQS, ∠PSQ = 90°. Given QS = 6 cm, calculate the length of AR.
In the given triangle PQR, LM is parallel to QR and PM : MR = 3 : 4.
Calculate the value of ratio:
- `(PL)/(PQ)` and then `(LM)/(QR)`
- `"Area of ΔLMN"/"Area of ΔMNR"`
- `"Area of ΔLQM"/"Area of ΔLQN"`
In the figure, given below, ABCD is a parallelogram. P is a point on BC such that BP : PC = 1 : 2. DP produced meets AB produces at Q. Given the area of triangle CPQ = 20 cm2.
Calculate:
- area of triangle CDP,
- area of parallelogram ABCD.
In the given figure, ∠B = ∠E, ∠ACD = ∠BCE, AB = 10.4 cm and DE = 7.8 cm. Find the ratio between areas of the ∆ABC and ∆DEC.
In the following diagram, lines l, m and n are parallel to each other. Two transversals p and q intersect the parallel lines at points A, B, C and P, Q, R as shown.
Prove that : `(AB)/(BC) = (PQ)/(QR)`
In the following figure, AD and CE are medians of ΔABC. DF is drawn parallel to CE. Prove that :
- EF = FB,
- AG : GD = 2 : 1
In the given figure, triangle ABC is similar to triangle PQR. AM and PN are altitudes whereas AX and PY are medians.
Prove that : `(AM)/(PN)=(AX)/(PY)`
In a triangle PQR, L and M are two points on the base QR, such that ∠LPQ = ∠QRP and ∠RPM = ∠RQP. Prove that:
- ΔPQL ∼ ΔRPM
- QL × RM = PL × PM
- PQ2 = QR × QL
In triangle ABC, AP : PB = 2 : 3. PO is parallel to BC and is extended to Q so that CQ is parallel to BA.
Find:
- area ΔAPO : area ΔABC.
- area ΔAPO : area ΔCQO.
