in triangle ABC if DE parallel To BC then prove that AD/AB=AE/AC
Answers
Step-by-step explanation:
:
ABC is a triangle and DE is parallel to BC.
:
AD/DB = AE/EC
:
* draw DM perpendicular to AC.
* And draw EN perpendicular to AB.
* Join BE and CD.
:
Area of triangle =1/2bh
In ADE =1/2 (AD)(EN),
In BDE =1/2 (DB) (EN).
Ar. Of ADE = 1/2 (AE)(DM)
Ar. Of Dec = 1/2 (EC) (DM).
.'. Ar.(ADE)/ Ar. (BDE)= 1/2AD* EN/ 1/2 DB* EN = AD / DB. - eq1⃣
Ar. ADE / are. DEC = 1/2 AE * DM/ 1/2 EC *DM = AC /EC. - eq✌️
BDE & DEC are lie on the same base and between the same parallels.
So, Ar. BDE = Ar. DEC
From eq one1⃣ and two✌️,
ᴀᴅ /ᴅʙ = ᴀᴇ /ᴇᴄ
Answer:
AD/AB=AE/A/C
ABC is a triangle and DE||BC
,AD/DB=AE/EC
- draw a DM perpendicular to AC
- EN parpendiculer to AB
- then join BE and CD
Area of triangle =1/2 bh
in triangle ADE=1/2 AD×EN
In triangle BDE =1/2 DB×EN
Ar. triangle ADE=1/2 AE×DM
At. triangle DEC=1/2 EC×DM
ar. triangle ADE/ ar. triangle BDE= 1/2 AD
EN=AD/DB -1
ar.of triangle ADE /ar. of triangle DEC = 1/2 AD
DM=AC/EC -2
BDE AND DEC are lie on same base and between same parallels
so ar of triangle BDE= ar of triangle DEC