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प्रश्न
The side AC of a triangle ABC is produced to point E so that CE = AC. D is the mid-point of BC and ED produced meets AB at F. Lines through D and C are drawn parallel to AB which meet AC at point P and EF at point R respectively.
Prove that:
- 3DF = EF
- 4CR = AB
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उत्तर
Consider the figure :
Here D is the midpoint of BC and DP is parallel to AB, therefore P is the midpoint of AC and
PD = `[1]/[2]`AB
(i) Again from the triangle AEF, we have AF || PD || CR and AP = `[1]/[3]`AE
Therefore, DF = `[1]/[3]` EF or we can say that 3DF = EF.
Hence, it is shown.
(ii) From the triangle PED, we have PD || CR and C is the midpoint of PE, therefore, CR = `[1]/[2]`PD
Now,
PD = `1/2` AB
`1/2"PD" = 1/4`AB
CR = `1/4`AB
4CR = AB
Hence, it is shown.
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संबंधित प्रश्न
ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that
- D is the mid-point of AC
- MD ⊥ AC
- CM = MA = `1/2AB`
L and M are the mid-point of sides AB and DC respectively of parallelogram ABCD. Prove that segments DL and BM trisect diagonal AC.
D and F are midpoints of sides AB and AC of a triangle ABC. A line through F and parallel to AB meets BC at point E.
- Prove that BDFE is a parallelogram
- Find AB, if EF = 4.8 cm.
In the figure, give below, 2AD = AB, P is mid-point of AB, Q is mid-point of DR and PR // BS. Prove that:
(i) AQ // BS
(ii) DS = 3 Rs.
In the given figure, AD and CE are medians and DF // CE.
Prove that: FB = `1/4` AB.
Prove that the straight lines joining the mid-points of the opposite sides of a quadrilateral bisect each other.
In ΔABC, the medians BE and CD are produced to the points P and Q respectively such that BE = EP and CD = DQ. Prove that: Q A and P are collinear.
In ΔABC, the medians BE and CD are produced to the points P and Q respectively such that BE = EP and CD = DQ. Prove that: A is the mid-point of PQ.
P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA of a quadrilateral ABCD in which AC = BD. Prove that PQRS is a rhombus.
E and F are respectively the mid-points of the non-parallel sides AD and BC of a trapezium ABCD. Prove that EF || AB and EF = `1/2` (AB + CD).
[Hint: Join BE and produce it to meet CD produced at G.]
