Answer:
Given: A Δ ABC in which , AB ≅ BC, BE − median of ΔABC, m∠ABE = 40°30'
To Find: ∠ABC, ∠ CE B,∠A E B
Solution: In Δ ABC , BE is the median.
So, AE= EC
Now, In Δ A E B and Δ CE B
AE = EC [ BE is median]
BE is common.
AB=BC [given]
Δ A E B ≅Δ CE B { SSS Congruency]⇒S→side
So, ∠ABE=∠C BE [ C PCT]
∴ ∠ABC=2×∠ABE=2×40°30'=81°
∠B EA =∠C E B [ C P CT]
Also,∠B A C = ∠BC A=k° [As AB =BC , if opposite sides are equal , then angle opposite to them are equal]
∠A+ ∠B+∠C=180°[ angle sum property of triangle]
k°+81°+k°=180°
2 k°=180°-81°
2 k°=99°
k°=99°/2
k°=49°30'[1°=60']
In Δ A E B, ∠ABE=40°30',∠A=49°30',∠A E B=?
∠ABE+∠A+∠A E B=180°[ angle sum property of triangle]
40°30'+49°30'+∠A E B=180°
90°+∠A E B=180°
∠A E B=180°-90°
∠A E B=90°
As, ∠B EA =∠C E B
So,∠C E B= 90°
So, BE is perpendicular bisector.
