Q. In ∆ABC.D is the midpoint of BC. E is the foot of the perpendicular from A to BC (lie on BC), and F is the foot of the perpendicular from D to AC. Given that BE =5, EC =9, and the area of triangle ABC is 84. Then the value of EF IS.
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Answers
Answer:
Let DE and DF be the perpendiculars from D on AB and AC respectively.
In △s BDE and CDF, DE=DF (Given)
∠BED=∠CFD=90
∘
BD=DC (∵ D is the mid-point of BC)
∴ △BDE≅△CDF (RHS)
⇒ ∠B=∠C (cpct)
⇒ AC=AB (Sides opp. equal ∠s are equal)
⇒ △ABC is isosceles.
Answer:
of side BC and ED is the bisector of ∠ADb, meeting AB in E, EF is drawn parallel to BC meeting AC in F.
To proof ∠EDF is a right angle.
Proof In △ADB, DE is the bisector of ∠ADB.
∴
DB
AD
=
EB
AE
⇒
DC
AD
=
EB
AE
.......(i) [∵ D is the mid-point of BC ∴ DB=DC]
In △ABC, we have
EF∣∣BC
⇒
DC
AD
=
FC
AF
⇒ In △ADC, DF divides AC in the ratio AD:DC
⇒ DF is the bisector of ∠ADC
Thus, DE and DF are the bisectors of adjacent supplementary angles ∠ADB and ∠ADCrespectively.
Hence, ∠EDF is a right angle.