Question

Hello !❤

Class 10th Maths question! ❤

In Triangle ABC, D is the midpoint of BC and DE is the bisector of Angle ADB and
EF || BC. Prove that Angle EDF is a right angle.

Do NoT sPaM !

Thank you for answering ❤​

Answer 1

In ∆ABC, we have EF∥BC.

So, AE/EB = AF/FC (Basic Proportionality Theorem) ....(1)

In ∆BDA, ED bisects ∠ADB, so ADBD = AEEB (Angle bisector theorem)

⇒AD/CD = AE/EB

[As, D is the mid point of BC, so BD = CD] ....(2)

from (1) and (2), we getAD/CD = AF/FC

Now, in ∆DAC, we have AD/CD = AF/FC (proved above)

⇒FD bisects ∠ADC [Converse of Angle bisector theorem]

⇒∠ADF = ∠FDC = 1/2∠ADC .....(3)

Since, ED bisects ∠ADB, then

∠EDA = ∠EDB = 1/2∠ADB ......(4)

Now, ∠ADB + ∠ADC = 180° [Linear pair]

⇒1/2∠ADB + 1/2∠ADC = 90°

⇒∠EDA + ∠ADF = 90° [Using (3) and (4)]

⇒∠EDF = 90°

⇒∠EDF is a right angle

HOPE IT HELPS!❤️

Answer 2

$\huge\underline\mathrm{GivEn:-}$

➡ A triangle ABC in which D is the mid-point of side BC and ED is the bisector of angle ADB, meeting AB in E. EF is drawn parallel to BC meeting AC in F.

$\huge\underline\mathrm{To Prove:-}$

➡ Angle EDF is a right angle.

$\huge\underline\mathrm{Proof:-}$

In triangle ABC, DE is the bisector of angle ADB.

➡ AD/DB = AE/EB

➡ AD/DC= AE/EB [ D is the mid point of BC i,e., DB = DC]__1)

In triangle ABC, we have

➡ EF || BC

AE/EB = AF/FC ___2)

➡ In triangle ABC, DF divides AC in the ratio AD:DC

➡ DF is the bisector of angle ADC

• Thus DE and DF are bisectors of adjacent supplementary angles angle ADB and angle ADC respectively.

• Hence, EDF is a right angle.

$\huge{\boxed{\boxed{\mathbb{\red{THANKS...}}}}}$