Advertisements
Advertisements
प्रश्न
ABCD is a parallelogram. The bisector of ∠BAD meets DC at P, and AD is half of AB.
Prove that: ∠APB is a right angle.
Advertisements
उत्तर
∠DAP =∠PAB
∠CBP = ∠PBA
∠DAB + ∠CBA = 180° ...(adjacent angles of || gm are supplementary)
Multipying by `(1)/(2)`
`(1)/(2)∠"DAB" + (1)/(1)∠"CBA" = (1)/(2) xx 180°`
∠PAB + ∠PBA = 90°
In ΔAPB,
∠PAB + ∠PBA + ∠APB = 180°
90° + ∠APB = 180°
∠APB = 90°
Therefore, ∠APB is a right angle.
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 diagonal BD of a parallelogram ABCD bisects angles B and D. Prove that ABCD is a rhombus.
In the alongside diagram, ABCD is a parallelogram in which AP bisects angle A and BQ bisects angle B.
Prove that:
- AQ = BP
- PQ = CD
- ABPQ 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.
In parallelogram ABCD, the bisector of angle A meets DC at P and AB = 2 AD.
Prove that:
(i) BP bisects angle B.
(ii) Angle APB = 90o.
Points M and N are taken on the diagonal AC of a parallelogram ABCD such that AM = CN. Prove that BMDN is a parallelogram.
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 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
