Question

Consider the following figure. ∆ABC is isosceles with AB = AC. DE is drawn parallel to BC. Give reasons for each of the following statements. i) angle B = angle C ii) angle ADE = angle B iii) angle AED = angle C iv) angle ADE = angle AED v) ∆ADE is isosceles.​

Answer 1

Answer:

Given : △ ABC is isosceles with AB=AC ,E and F are the mid-points of BC, CA and AB

To prove: AD⊥EFand is bisected by t

construction: Join D, F and F

Proof: DE∣∣AC and DE= 1/2AB

and DF∣∣Ac And DE= 1/2AC

The line segment joining midpoints of two sides of a triangle is parallel to the third side and is half of it

DE = DF (∵AB=AC) Also AF=AE

∴AF= 1/2

AB,AE= 1/2 AC

∴DE=AE=AF=DF

and also DF∣∣ AE and DE∣∣AF

⇒ DEAF is a rhombus.

since diagrams of a rhombus bisect each other of right angles

∴AD⊥EF and is bisected by it

Step-by-step explanation:

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