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

The median of the isosceles triangle ABC bisects the side AC, forming two

congruent triangles ΔBAE and ΔBEC.

Reasons:

Given parameters are;

In ΔABC

AB is congruent to BC; AB ≅ BC

A median of ΔABC = BE

m∠ABE = 40°30'

Required:

m∠ABC

Solution:

AB ≅ BC

∴ AB = BC,  ∴ ΔABC is an isosceles triangle

∠BAE = ∠BCE base angles of an isosceles triangle ΔABC

Line BE bisects AC; definition of median line

AE = EC definition of bisected line

ΔBAE ≅ ΔBEC by Side-Angle-Side rule of congruency

∠ABE = ∠CBE by Congruent Parts of Congruent Triangles are Congruent, CPCTC

∠ABC = ∠ABE + ∠CBE by angle addition postulate

∴ ∠ABC = ∠ABE + ∠ABE = 2× ∠ABE by substitution property

m∠ABC = 2 × 40°30' = 81°

∠FEC = ∠BEA by vertical angle theorem

∠BEA = ∠BEC by CPCTC

∠BEA + ∠BEC = 180° by linear pair angles

∴ 2 × ∠BEA = 180°

m∠FEC = m∠BEA = 90°

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step-by-step explanation: