ABC is an equilateral triangle.
P. is any point within the trian-
gle. AD I BC. PX, PY and
PZ are perpendiculars on sides
BC, AC and AB. If height of
the AABC, AD is 3.5 cm, find
the sum of PX, PY, PZ.
Answers
Step-by-step explanation:
∠ CBP = 30°, \angle∠ BCP = 60° with BP>PC
Step-by-step explanation:
Given, ABC is an equilateral triangle.
With all angles equal,
\angle A = \angle B = \angle C = 60∠A=∠B=∠C=60
In ABP, Clearly, \angle A > \angle ABP∠A>∠ABP
So, BP > PA Proved!
In BPC,
\angle C > \angle CBP∠C>∠CBP
So, BP > PC Proved!
Let P point in AC be such that it bisects AC
So, BP is the median on AC of equilateral triangle ABC.
A median bisects the angles in equilateral and isosceles triangle
\angle ABP = \angle CBP∠ABP=∠CBP
So, AP = CP
\angle A = \angle B = \angle C = 60∠A=∠B=∠C=60 (ABC is equilateral)
\angle ABP = \angle CBP = \frac {60}{2} = 30∠ABP=∠CBP=
2
60
=30
\angle BAP > \angle ABP∠BAP>∠ABP
BP > PA (Sides opposite to greater angles are greater) Proved!
BP > PC
Step-by-step explanation:
From ΔABM,
Let AM⊥BC
AB
2
=BM
2
+AM
2
=(
2
3x
)
2
+AM
2
=
4
9x
2
+(AP
2
−MP
2
) ...[from ΔAMP]
=
4
9x
2
+AP
2
−(
2
x
)
2
AB
2
=AP
2
+2x
2
AB
2
−2x
2
=AP
2
9AB
2
−18x
2
=9AP
2
....[multiplying by 9]
9AB
2
−2(9x
2
)=9AP
2
9AB
2
−2(3x)
2
=9AP
2
9AB
2
−2(AB)
2
=9AP
2
7AB
2
=9AP
2
Hence proved.