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°

Learn more here:

link

Since we want to solve for the variable x, we want to isolate x

a²x + (a - 1) = (a + 1)x ⇒ Distribute x to (a+1). Also, remove parentheses

a²x + a - 1 = ax + x ⇒ Subtract a from both sides

a²x - 1 = ax + x - a ⇒ Add 1 to both sides

a²x = ax + x - a + 1 ⇒ Subtract (ax + x) from both sides

a²x - (ax + x)= ax + x - a + 1 - (ax+x) ⇒ Simplify. Remember that multiplying positive by negative = negative

a²x - ax - x = ax + x - a + 1 - ax - x ⇒ Simplify

a²x - ax - x = -a + 1 ⇒ Factor out the x from a²x - ax - x

x(a² - a - 1) = -a + 1 ⇒ Divide both sides by (a² - a - 1)

x = (-a + 1) / (a² - a - 1)

However, we need to make sure that the denominator does not equal 0. Therefore, you set the denominator = 0 (just use the quadratic formula for this), and it gives that the denominator =0 when a = (1+√5)/2 AND (1-√5)/2

Therefore, the final answer is

x = (-a + 1) / (a² - a - 1) given that a ≠ (1+√5)/2, a ≠ (1-√5)/2