Question

bisector of angle BAC of triangle ABC and the straight line through the mid point D of the side AC and parallel to the side AB meet at a point E outside BC. Prove that angle AEC is equal to 1 right angle ​

Answer 1

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Asked on November 22, 2019 byAnshul Ghosh

In △ABC, D is the mid-point of BC and ED is the bisector of the ∠ADB and EF is drawn parallel to BC cutting AC at F. Prove that ∠EDF is a right angle.



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ANSWER

Given A △ABC in which D is the mid-point 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.

∴    DBAD=EBAE

⇒  DCAD=EBAE.......(i)         [∵ D is the mid-point of BC ∴ DB=DC]

In △ABC, we have

         EF∣∣BC

⇒   DCAD=FCAF

⇒   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.

Answer 2

AB//DE and AE is the transversal

<BAE= alternate <AED = x (let)

<BAE=<CAE [ AE is the bisector of <BAC]

<BAE=<CAE=<AED=x

<AED=<CAE

AD=DE [ side opposite to equal angles are equal]

Since, D is the mid point of AC

AD=CD

AD=CD=DE

<AED=<DEC=<DCE

we know that,

<EDC=<DAE + <AED = x+x = 2x

In triangle DEC,

<DEC + <DCE + <EDC = 180

x + x + 2x = 180

4x = 180

x = 180/4

x = 45

<AEC= 45 + 45 = 90

hope my answer helps you!