Advertisements
Advertisements
प्रश्न
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°
Advertisements
उत्तर
Diagonals of a kite intersect at right angles
∴ ∠MON = 90° .......(i)
In ΔBCD,
E and F are mid-points of CD and BC respectively.
Therefore, EF || DB and EF = `(1)/(2)"DB"` .......(ii)
EF || DB ⇒ MF || ON
∴ ∠MON + ∠MFN = 180°
⇒ 90° + ∠MFN = 180°
⇒ ∠MFN = 90°
⇒ ∠EFG = 90°.
APPEARS IN
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`
Let Abc Be an Isosceles Triangle in Which Ab = Ac. If D, E, F Be the Mid-points of the Sides Bc, Ca and a B Respectively, Show that the Segment Ad and Ef Bisect Each Other at Right Angles.
In the given figure, seg PD is a median of ΔPQR. Point T is the mid point of seg PD. Produced QT intersects PR at M. Show that `"PM"/"PR" = 1/3`.
[Hint: DN || QM]
D, E, and F are the mid-points of the sides AB, BC and CA of an isosceles ΔABC in which AB = BC.
Prove that ΔDEF is also isosceles.
The following figure shows a trapezium ABCD in which AB // DC. P is the mid-point of AD and PR // AB. Prove that:
PR = `[1]/[2]` ( AB + CD)
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.
In trapezium ABCD, sides AB and DC are parallel to each other. E is mid-point of AD and F is mid-point of BC.
Prove that: AB + DC = 2EF.
In the given figure, ABCD is a trapezium. P and Q are the midpoints of non-parallel side AD and BC respectively. Find: PQ, if AB = 12 cm and DC = 10 cm.
In the given figure, PS = 3RS. M is the midpoint of QR. If TR || MN || QP, then prove that:
RT = `(1)/(3)"PQ"`
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.
