Advertisements
Advertisements
प्रश्न
In below fig. ABCD is a parallelogram and E is the mid-point of side B If DE and AB when produced meet at F, prove that AF = 2AB.
उत्तर
In Δ BEF and ΔCED
`∠`BEF = `∠`CED [Verified opposite angle]
BE = CE [ ∵ E is the mid-point of BC]
`∠`EBF = `∠`ECD [∵ Alternate interior angles are equal]
∴ ∇ BEF ≅ Δ CED [Angle side angle congruence]
∴ BF = CD [Corresponding Parts of Congruent Triangles]
AF = AB + AF
AF = AB + AB
AF = 2 AB
APPEARS IN
संबंधित प्रश्न
ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see the given figure). AC is a diagonal. Show that:
- SR || AC and SR = `1/2AC`
- PQ = SR
- PQRS is a parallelogram.
In the given figure, `square`PQRS and `square`MNRL are rectangles. If point M is the midpoint of side PR then prove that,
- SL = LR
- LN = `1/2`SQ
In the adjacent figure, `square`ABCD is a trapezium AB || DC. Points M and N are midpoints of diagonal AC and DB respectively then prove that MN || AB.
In triangle ABC, the medians BP and CQ are produced up to points M and N respectively such that BP = PM and CQ = QN. Prove that:
- M, A, and N are collinear.
- A is the mid-point of MN.
If the quadrilateral formed by joining the mid-points of the adjacent sides of quadrilateral ABCD is a rectangle,
show that the diagonals AC and BD intersect at the right angle.
Show that the quadrilateral formed by joining the mid-points of the adjacent sides of a square is also a square.
ABCD is a kite in which BC = CD, AB = AD. E, F and G are the mid-points of CD, BC and AB respectively. Prove that: ∠EFG = 90°
E is the mid-point of the side AD of the trapezium ABCD with AB || DC. A line through E drawn parallel to AB intersect BC at F. Show that F is the mid-point of BC. [Hint: Join AC]
P, Q, R and S are respectively the mid-points of sides AB, BC, CD and DA of quadrilateral ABCD in which AC = BD and AC ⊥ BD. Prove that PQRS is a square.
P and Q are the mid-points of the opposite sides AB and CD of a parallelogram ABCD. AQ intersects DP at S and BQ intersects CP at R. Show that PRQS is a parallelogram.
