ABCD is a parallelogram in which angle dab=80.bisector of angle a and b meets CD at p prove that:ad=dp,cp=cab,dc=2ad
Answers
Answer:
∞
Step-by-step explanation:
ABCD is a parallelogram, in which ∠A = 80°
⇒ ∠B = 80° [Adjacent angles of a parallelogram are supplementary]
∠C = 80° = ∠A [Opposite angles of a parallelogram are equal]
∠D = ∠B = 80° [Opposite angles of parallelogram are equal]
So, here it is concluded that ∠A=∠B=∠C=∠D=80°
If all the 4 angles of the parallelogram are 80° then it is not possible.
Because,
80°+80°+80°+80°=320°
But we know that the angle sum of a quadrilateral is always 360°.
So, such a parallelogram is not possible.
Answer:
It is easy.
Step-by-step explanation:
Angle B = 180-80=100⁰. This is because adjacent angles are supplementary.
Angle D= 100⁰ because opposite angles equal.
Angle C=80⁰ because opposite angles equal.
Angle PAB= 40⁰= Angle DAP because AP is a bisector.
Angle PBA=50⁰=Angle PBC because BP is a bisector.
Angle DPA=180-(100+40)=40⁰ by Angle Sum Property.
Angle BPC=180-(80+50)=50⁰ by Angle Sum Property.
In Triangle DPA,
ANGLE DAP=ANGLE DPA.
THEREFORE, TRIANGLE IS ISOSCELES.
AND SO
(i). AD=DP
In Triangle BPC,
ANGLE CBP=ANGLE BPC.
THEREFORE, TRIANGLE IS ISOSCELES.
AND SO
(ii). CP=BC
AD=BC because opposite sides in a parallelogram are always equal.
AD=DP (PROVEN)
PC=BC(PROVEN)
SO,
LHS=DC
DC=DP+PC
=AD+BC
=AD+AD (AD=BC)
=2AD
=RHS.
THEREFORE HENCE PROVEN.
IT WAS ASKED IN THE ICSE CLASS 9 JANUARY BOARDS IN QUESTION 11 SUB-PART A FOR 3 MARKS.
I HAD GOTTEN 78½ ON 80 IN THAT MATHS PAPER