Advertisements
Advertisements
प्रश्न
Prove that the bisectors of opposite angles of a parallelogram are parallel.
Advertisements
उत्तर
Given ABCD is a parallelogram. The bisectors of ∠ADC and ∠BCD meet at E. The bisectors of ∠ABC and ∠BCD meet at F
From the parallelogram ABCD we have
∠ADC + ∠BCD = 180° ...[ sum of adjacent angles of a parallalogram ]
⇒ `("∠ADC")/(2) + ("∠BCD")/(2)` = 90°
⇒ ∠EDC + ∠EDC = 90°
In triangle ECD sum of angles = 180°
⇒ ∠EDC + ∠ECD + ∠CED = 180°
= ∠CED = 90°
Similarly taking triangle BCF it can prove that ∠BFC = 90°
Now since
∠BFC = ∠CED = 90°
Therefore the lines DE and BF are parallel
Hence proved
APPEARS IN
संबंधित प्रश्न
E is the mid-point of side AB and F is the mid-point of side DC of parallelogram ABCD. Prove that AEFD is a parallelogram.
The following figure shows a trapezium ABCD in which AB is parallel to DC and AD = BC. 
Prove that:
(i) ∠DAB = ∠CBA
(ii) ∠ADC = ∠BCD
(iii) AC = BD
(iv) OA = OB and OC = OD.
PQRS is a parallelogram. T is the mid-point of PQ and ST bisects ∠PSR.
Prove that: QR = QT
PQRS is a parallelogram. T is the mid-point of PQ and ST bisects ∠PSR.
Prove that: ∠RTS = 90°
ABCD is a parallelogram. The bisector of ∠BAD meets DC at P, and AD is half of AB.
Prove that: BP bisects ∠ABC.
In a parallelogram ABCD, E is the midpoint of AB and DE bisects angle D. Prove that: BC = BE.
In the Figure, ABCD is a rectangle and EFGH is a parallelogram. Using the measurements given in the figure, what is the length d of the segment that is perpendicular to `bar("HE")` and `bar("FG")`?
In parallelogram ABCD of the accompanying diagram, line DP is drawn bisecting BC at N and meeting AB (extended) at P. From vertex C, line CQ is drawn bisecting side AD at M and meeting AB (extended) at Q. Lines DP and CQ meet at O. Show that the area of triangle QPO is `9/8` of the area of the parallelogram ABCD
In the following figure, ABCD and AEFG are two parallelograms. If ∠C = 55º, determine ∠F.
Construct a parallelogram POUR in which, PO = 5.5 cm, OU = 7.2 cm and ∠O = 70°.
