In a figure AD is the median of a triangle abc and AL transversal to BC then prove that area of ABD equal to area of ACD equal to half of area of ABC
Answers
Answer:
Step-by-step explanation:
Given,
AD is the median
So,
BD=DC,
<BAD=<DAC
To prove :ar(ABD)=ar(CAD)
Proof,
In ABD and CAD,
BD=DC(given)
<BAD=<CAD(given)
AD=AD(given)
Thus,ABD=~CAD
So,ar(ABD)=ar(CAD)
=1/2ar(ABC)
(a) Given:
∆ABC in which AD is the median. P is any point on AD. Join PB and PC.
To prove:
(i) Area of ∆PBD = area of ∆PDC.
(ii) Area of ∆ABP = area of ∆ACP.
Proof:
From fig (1)
AD is a median of ∆ABC
So, ar (∆ABD) = ar (∆ADC) …. (1)
Also, PD is the median of ∆BPD
Similarly, ar (∆PBD) = ar (∆PDC) …. (2)
Now, let us subtract (2) from (1), we get
ar (∆ABD) – ar (∆PBD) = ar (∆ADC) – ar (∆PDC)
Or ar (∆ABP) = ar (∆ACP)
Hence proved.
(b) Given:
∆ABC in which DE || BC
To prove:
(i) area of ∆ACD = area of ∆ ABE.
(ii) Area of ∆OBD = area of ∆OCE.
Proof:
From fig (2)
∆DEC and ∆BDE are on the same base DE and between the same || line DE and BE.
ar (∆DEC) = ar (∆BDE)
Now add ar (ADE) on both sides, we get
ar (∆DEC) + ar (∆ADE) = ar (∆BDE) + ar (∆ADE)
ar (∆ACD) = ar (∆ABE)
Hence proved.
Similarly, ar (∆DEC) = ar (∆BDE)
Subtract ar (∆DOE) from both sides, we get
ar (∆DEC) – ar (∆DOE) = ar (∆BDE) – ar (∆DOE)
ar (∆OBD) = ar (∆OCE)
Hence proved.