Chapter 5.6, Problem 39E

### Elementary Geometry For College St...

7th Edition
Alexander + 2 others
ISBN: 9781337614085

Chapter
Section

### Elementary Geometry For College St...

7th Edition
Alexander + 2 others
ISBN: 9781337614085
Textbook Problem

# Given: Δ A B C (not shown) is isosceles with m ∠ A B C = m ∠ C = 72 ∘ ; B D → bisects ∠ A B C and A B = 1 Find: B C

To determine

To find:

BC.

Explanation

Given:

In an isosceles Î”ABC with mâˆ ABC=mâˆ C=72âˆ˜, BDâ†’ bisects âˆ ABC and AB=1.

Theorem used:

Angle bisector theorem:

If a ray bisects one angle of a triangle, then it divides the opposite side into segments whose lengths are proportional to the lengths of the two sides that form the bisected angle.

Calculation:

Consider an isosceles Î”ABC.

We have mâˆ ABC=mâˆ C=72âˆ˜.

Since BDâ†’ bisects âˆ ABC, mâˆ ABD=mâˆ DBC=12âˆ ABC.

mâˆ ABD=12âˆ ABC=12(72âˆ˜)=36âˆ˜

We know that the sum of angles in a triangle is 180âˆ˜.

mâˆ A+mâˆ B+mâˆ C=180âˆ˜mâˆ A+72âˆ˜+72âˆ˜=180âˆ˜mâˆ A+144âˆ˜=180âˆ˜mâˆ A=180âˆ˜âˆ’144âˆ˜mâˆ A=36âˆ˜

Thus mâˆ ABD=mâˆ A=36âˆ˜

â‡’Î”ABD is an isosceles triangle, with AD=DB.

Consider the Î”BDC.

mâˆ BDC+mâˆ DBC+mâˆ DCB=180âˆ˜mâˆ BDC+36âˆ˜+72âˆ˜=180âˆ˜mâˆ BDC+108âˆ˜=180âˆ˜mâˆ BDC=180âˆ˜âˆ’108âˆ˜mâˆ BDC=72âˆ˜

That is, mâˆ BDC=mâˆ DCB

Hence Î”BDC is an isosceles triangle with BD=BC

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