in triangle ABC, D is the mid point of side AC such that BD =half AC. show that angle ABC is a right angle.
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In ΔADB, AD = BD
∠DAB = ∠DBA = ∠x ( these are the angles which have opposite sides)
In ΔDCB, BD = CD
∠DBC = ∠DCB = ∠y
In ΔABC we will use the angle sum property
∠ABC + ∠BCA + ∠CAB = 180°
2(∠x + ∠y) = 180°
∠x + ∠y = 90°
∠ABC = 90°
This means that ABC is the right angled triangle.
Hope it helps you!!
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