Given: δabc, ab ≅ bc, be − median of δabc, m∠abe = 40°30' find: m∠abc, m∠fec

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.

could you provide figure.