In triangle ABC, X the midpoint of AC, and AD bisects angle BAC, meeting BC in D. XE is drawn parallel to AB and meets AD in E. Prove that angle AEC =90°
Answers
Answer:
Given: ΔABC, the bisector of ∠BAC, D is the midpoint of side AC, line parallel to side AB through point D, the bisector and parallel line meet at point E outside BC
To prove ∠AEC = 1 right angle, i.e., ∠AEC = 90°
In ΔABC, let AE be the bisector of ∠BAC,
Let DE be the parallel line to side AB though point D, where D is the midpoint of AC.
So the figure of the given question is as shown below,
Given AE is the bisector of ∠BAC, so let
∠BAE = ∠EAC = x……….(i)
Given DE||AB and with AE as the transversal line,
Then ∠BAE = ∠AED = x ……..(ii) (as they form alternate interior angles)
Now consider ΔADE,
From equation (ii),
∠BAE = ∠AED = x
We know sides opposite to equal angles are equal, so
AD = DE………(iii)
And in same triangle ∠EDC form external angle, but we know the measure of any exterior angle of a triangle is equal to the sum of the measures of its two interior opposite angles, so
∠EDC = ∠EAD + ∠AED
Now substituting values from equation (i) and (ii), we get
∠EDC = x + x = 2x….(iv)
Now consider ΔAEC, given D is midpoint of AC,
So AD = DC……(iv)
We know angles opposite to equal sides are equal, so
∠AED = ∠DEC
Now substituting the value from equation (ii), we get
∠AED = DEC = x………(v)
By comparing equation (iii) and (iv), we get
DE = DC…..(v)
Now consider ΔDEC
We also know angles opposite to equal sides are equal, so from equation (v), we get
∠DEC = ∠ECD
Now substituting the value from equation (ii), we get
∠DEC = ∠ECD = x……….(vi)
Now in ΔEDC,
We know in a triangle the sum of all three interior angles is equal to 180°.
So in this case,
∠EDC + ∠DEC + ∠ECD = 180°
Substituting value from equation (iv) and (vi) in above equation we get
2x + x + x = 180°
⇒ 4x = 180°
⇒ x = 45°……….(vii)
Now from figure,
∠AEC = ∠AED + ∠DEC
Substituting values from equation (v), we get
∠AEC = x + x = 2x
Substituting values from equation (vii), we get
∠AEC = 2(45°) = 90°
Hence ∠AEC = 1 right angle
Hence proved